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List of trigonometric identities

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Intrigonometry,trigonometric identities areequalities that involvetrigonometric functions and are true for every value of the occurringvariables for which both sides of the equality are defined. Geometrically, these areidentities involving certain functions of one or moreangles. They are distinct fromtriangle identities, which are identities potentially involving angles but also involving side lengths or other lengths of atriangle.

These identities are useful whenever expressions involving trigonometric functions need to be simplified. An important application is theintegration of non-trigonometric functions: a common technique involves first using thesubstitution rule with a trigonometric function, and then simplifying the resulting integral with a trigonometric identity.

Pythagorean identities

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Main article:Pythagorean trigonometric identity
Trigonometric functions and their reciprocals on the unit circle. All of the right-angled triangles are similar, i.e. the ratios between their corresponding sides are the same. For sin, cos and tan the unit-length radius forms the hypotenuse of the triangle that defines them. The reciprocal identities arise as ratios of sides in the triangles where this unit line is no longer the hypotenuse. The triangle shaded blue illustrates the identity1+cot2⁡θ=csc2⁡θ{\displaystyle 1+\cot ^{2}\theta =\csc ^{2}\theta }, and the red triangle shows thattan2⁡θ+1=sec2⁡θ{\displaystyle \tan ^{2}\theta +1=\sec ^{2}\theta }.

The basic relationship between thesine and cosine is given by the Pythagorean identity:

sin2⁡θ+cos2⁡θ=1,{\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1,}

wheresin2⁡θ{\displaystyle \sin ^{2}\theta } means(sin⁡θ)2{\displaystyle {(\sin \theta )}^{2}} andcos2⁡θ{\displaystyle \cos ^{2}\theta } means(cos⁡θ)2.{\displaystyle {(\cos \theta )}^{2}.}

This can be viewed as a version of thePythagorean theorem, and follows from the equationx2+y2=1{\displaystyle x^{2}+y^{2}=1} for theunit circle. This equation can be solved for either the sine or the cosine:

sin⁡θ=±1−cos2⁡θ,cos⁡θ=±1−sin2⁡θ.{\displaystyle {\begin{aligned}\sin \theta &=\pm {\sqrt {1-\cos ^{2}\theta }},\\\cos \theta &=\pm {\sqrt {1-\sin ^{2}\theta }}.\end{aligned}}}

where the sign depends on thequadrant ofθ.{\displaystyle \theta .}

Dividing this identity bysin2⁡θ{\displaystyle \sin ^{2}\theta },cos2⁡θ{\displaystyle \cos ^{2}\theta }, or both yields the following identities:1+cot2⁡θ=csc2⁡θ1+tan2⁡θ=sec2⁡θsec2⁡θ+csc2⁡θ=sec2⁡θcsc2⁡θ{\displaystyle {\begin{aligned}&1+\cot ^{2}\theta =\csc ^{2}\theta \\&1+\tan ^{2}\theta =\sec ^{2}\theta \\&\sec ^{2}\theta +\csc ^{2}\theta =\sec ^{2}\theta \csc ^{2}\theta \end{aligned}}}

Using these identities, it is possible to express any trigonometric function in terms of any other (up to a plus or minus sign):

Each trigonometric function in terms of each of the other five.[1]
in terms ofsin⁡θ{\displaystyle \sin \theta }csc⁡θ{\displaystyle \csc \theta }cos⁡θ{\displaystyle \cos \theta }sec⁡θ{\displaystyle \sec \theta }tan⁡θ{\displaystyle \tan \theta }cot⁡θ{\displaystyle \cot \theta }
sin⁡θ={\displaystyle \sin \theta =}sin⁡θ{\displaystyle \sin \theta }1csc⁡θ{\displaystyle {\frac {1}{\csc \theta }}}±1−cos2⁡θ{\displaystyle \pm {\sqrt {1-\cos ^{2}\theta }}}±sec2⁡θ−1sec⁡θ{\displaystyle \pm {\frac {\sqrt {\sec ^{2}\theta -1}}{\sec \theta }}}±tan⁡θ1+tan2⁡θ{\displaystyle \pm {\frac {\tan \theta }{\sqrt {1+\tan ^{2}\theta }}}}±11+cot2⁡θ{\displaystyle \pm {\frac {1}{\sqrt {1+\cot ^{2}\theta }}}}
csc⁡θ={\displaystyle \csc \theta =}1sin⁡θ{\displaystyle {\frac {1}{\sin \theta }}}csc⁡θ{\displaystyle \csc \theta }±11−cos2⁡θ{\displaystyle \pm {\frac {1}{\sqrt {1-\cos ^{2}\theta }}}}±sec⁡θsec2⁡θ−1{\displaystyle \pm {\frac {\sec \theta }{\sqrt {\sec ^{2}\theta -1}}}}±1+tan2⁡θtan⁡θ{\displaystyle \pm {\frac {\sqrt {1+\tan ^{2}\theta }}{\tan \theta }}}±1+cot2⁡θ{\displaystyle \pm {\sqrt {1+\cot ^{2}\theta }}}
cos⁡θ={\displaystyle \cos \theta =}±1−sin2⁡θ{\displaystyle \pm {\sqrt {1-\sin ^{2}\theta }}}±csc2⁡θ−1csc⁡θ{\displaystyle \pm {\frac {\sqrt {\csc ^{2}\theta -1}}{\csc \theta }}}cos⁡θ{\displaystyle \cos \theta }1sec⁡θ{\displaystyle {\frac {1}{\sec \theta }}}±11+tan2⁡θ{\displaystyle \pm {\frac {1}{\sqrt {1+\tan ^{2}\theta }}}}±cot⁡θ1+cot2⁡θ{\displaystyle \pm {\frac {\cot \theta }{\sqrt {1+\cot ^{2}\theta }}}}
sec⁡θ={\displaystyle \sec \theta =}±11−sin2⁡θ{\displaystyle \pm {\frac {1}{\sqrt {1-\sin ^{2}\theta }}}}±csc⁡θcsc2⁡θ−1{\displaystyle \pm {\frac {\csc \theta }{\sqrt {\csc ^{2}\theta -1}}}}1cos⁡θ{\displaystyle {\frac {1}{\cos \theta }}}sec⁡θ{\displaystyle \sec \theta }±1+tan2⁡θ{\displaystyle \pm {\sqrt {1+\tan ^{2}\theta }}}±1+cot2⁡θcot⁡θ{\displaystyle \pm {\frac {\sqrt {1+\cot ^{2}\theta }}{\cot \theta }}}
tan⁡θ={\displaystyle \tan \theta =}±sin⁡θ1−sin2⁡θ{\displaystyle \pm {\frac {\sin \theta }{\sqrt {1-\sin ^{2}\theta }}}}±1csc2⁡θ−1{\displaystyle \pm {\frac {1}{\sqrt {\csc ^{2}\theta -1}}}}±1−cos2⁡θcos⁡θ{\displaystyle \pm {\frac {\sqrt {1-\cos ^{2}\theta }}{\cos \theta }}}±sec2⁡θ−1{\displaystyle \pm {\sqrt {\sec ^{2}\theta -1}}}tan⁡θ{\displaystyle \tan \theta }1cot⁡θ{\displaystyle {\frac {1}{\cot \theta }}}
cot⁡θ={\displaystyle \cot \theta =}±1−sin2⁡θsin⁡θ{\displaystyle \pm {\frac {\sqrt {1-\sin ^{2}\theta }}{\sin \theta }}}±csc2⁡θ−1{\displaystyle \pm {\sqrt {\csc ^{2}\theta -1}}}±cos⁡θ1−cos2⁡θ{\displaystyle \pm {\frac {\cos \theta }{\sqrt {1-\cos ^{2}\theta }}}}±1sec2⁡θ−1{\displaystyle \pm {\frac {1}{\sqrt {\sec ^{2}\theta -1}}}}1tan⁡θ{\displaystyle {\frac {1}{\tan \theta }}}cot⁡θ{\displaystyle \cot \theta }


Reflections, shifts, and periodicity

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By examining the unit circle, one can establish the following properties of the trigonometric functions.

Reflections

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Unit circle with a swept angle theta plotted at coordinates (a,b). As the angle is reflected in increments of one-quarter pi, or 45 degrees, the coordinates are transformed. For a transformation of one-quarter pi (45 degrees, or 90 – theta), the coordinates are transformed to (b,a). Another increment of the angle of reflection by one-quarter pi (90 degrees total, or 180 – theta) transforms the coordinates to (-a,b). A third increment of the angle of reflection by another one-quarter pi (135 degrees total, or 270 – theta) transforms the coordinates to (-b,-a). A final increment of one-quarter pi (180 degrees total, or 360 – theta) transforms the coordinates to (a,-b).
Transformation of coordinates (a,b) when shifting the reflection angleα{\displaystyle \alpha } in increments ofπ4{\displaystyle {\frac {\pi }{4}}}

When the direction of aEuclidean vector is represented by an angleθ,{\displaystyle \theta ,} this is the angle determined by the free vector (starting at the origin) and the positivex{\displaystyle x}-unit vector. The same concept may also be applied to lines in anEuclidean space, where the angle is that determined by a parallel to the given line through the origin and the positivex{\displaystyle x}-axis. If a line (vector) with directionθ{\displaystyle \theta } is reflected about a line with directionα,{\displaystyle \alpha ,} then the direction angleθ′{\displaystyle \theta ^{\prime }} of this reflected line (vector) has the valueθ′=2α−θ.{\displaystyle \theta ^{\prime }=2\alpha -\theta .}

The values of the trigonometric functions of these anglesθ,θ′{\displaystyle \theta ,\;\theta ^{\prime }} for specific anglesα{\displaystyle \alpha } satisfy simple identities: either they are equal, or have opposite signs, or employ the complementary trigonometric function. These are also known asreduction formulae.[2]

θ{\displaystyle \theta } reflected inα=0{\displaystyle \alpha =0}[3]
odd/even identities		θ{\displaystyle \theta } reflected inα=π4{\displaystyle \alpha ={\frac {\pi }{4}}}θ{\displaystyle \theta } reflected inα=π2{\displaystyle \alpha ={\frac {\pi }{2}}}θ{\displaystyle \theta } reflected inα=3π4{\displaystyle \alpha ={\frac {3\pi }{4}}}θ{\displaystyle \theta } reflected inα=π{\displaystyle \alpha =\pi }
compare toα=0{\displaystyle \alpha =0}
sin⁡(−θ)=−sin⁡θ{\displaystyle \sin(-\theta )=-\sin \theta }sin⁡(π2−θ)=cos⁡θ{\displaystyle \sin \left({\tfrac {\pi }{2}}-\theta \right)=\cos \theta }sin⁡(π−θ)=+sin⁡θ{\displaystyle \sin(\pi -\theta )=+\sin \theta }sin⁡(3π2−θ)=−cos⁡θ{\displaystyle \sin \left({\tfrac {3\pi }{2}}-\theta \right)=-\cos \theta }sin⁡(2π−θ)=−sin⁡(θ)=sin⁡(−θ){\displaystyle \sin(2\pi -\theta )=-\sin(\theta )=\sin(-\theta )}
cos⁡(−θ)=+cos⁡θ{\displaystyle \cos(-\theta )=+\cos \theta }cos⁡(π2−θ)=sin⁡θ{\displaystyle \cos \left({\tfrac {\pi }{2}}-\theta \right)=\sin \theta }cos⁡(π−θ)=−cos⁡θ{\displaystyle \cos(\pi -\theta )=-\cos \theta }cos⁡(3π2−θ)=−sin⁡θ{\displaystyle \cos \left({\tfrac {3\pi }{2}}-\theta \right)=-\sin \theta }cos⁡(2π−θ)=+cos⁡(θ)=cos⁡(−θ){\displaystyle \cos(2\pi -\theta )=+\cos(\theta )=\cos(-\theta )}
tan⁡(−θ)=−tan⁡θ{\displaystyle \tan(-\theta )=-\tan \theta }tan⁡(π2−θ)=cot⁡θ{\displaystyle \tan \left({\tfrac {\pi }{2}}-\theta \right)=\cot \theta }tan⁡(π−θ)=−tan⁡θ{\displaystyle \tan(\pi -\theta )=-\tan \theta }tan⁡(3π2−θ)=+cot⁡θ{\displaystyle \tan \left({\tfrac {3\pi }{2}}-\theta \right)=+\cot \theta }tan⁡(2π−θ)=−tan⁡(θ)=tan⁡(−θ){\displaystyle \tan(2\pi -\theta )=-\tan(\theta )=\tan(-\theta )}
csc⁡(−θ)=−csc⁡θ{\displaystyle \csc(-\theta )=-\csc \theta }csc⁡(π2−θ)=sec⁡θ{\displaystyle \csc \left({\tfrac {\pi }{2}}-\theta \right)=\sec \theta }csc⁡(π−θ)=+csc⁡θ{\displaystyle \csc(\pi -\theta )=+\csc \theta }csc⁡(3π2−θ)=−sec⁡θ{\displaystyle \csc \left({\tfrac {3\pi }{2}}-\theta \right)=-\sec \theta }csc⁡(2π−θ)=−csc⁡(θ)=csc⁡(−θ){\displaystyle \csc(2\pi -\theta )=-\csc(\theta )=\csc(-\theta )}
sec⁡(−θ)=+sec⁡θ{\displaystyle \sec(-\theta )=+\sec \theta }sec⁡(π2−θ)=csc⁡θ{\displaystyle \sec \left({\tfrac {\pi }{2}}-\theta \right)=\csc \theta }sec⁡(π−θ)=−sec⁡θ{\displaystyle \sec(\pi -\theta )=-\sec \theta }sec⁡(3π2−θ)=−csc⁡θ{\displaystyle \sec \left({\tfrac {3\pi }{2}}-\theta \right)=-\csc \theta }sec⁡(2π−θ)=+sec⁡(θ)=sec⁡(−θ){\displaystyle \sec(2\pi -\theta )=+\sec(\theta )=\sec(-\theta )}
cot⁡(−θ)=−cot⁡θ{\displaystyle \cot(-\theta )=-\cot \theta }cot⁡(π2−θ)=tan⁡θ{\displaystyle \cot \left({\tfrac {\pi }{2}}-\theta \right)=\tan \theta }cot⁡(π−θ)=−cot⁡θ{\displaystyle \cot(\pi -\theta )=-\cot \theta }cot⁡(3π2−θ)=+tan⁡θ{\displaystyle \cot \left({\tfrac {3\pi }{2}}-\theta \right)=+\tan \theta }cot⁡(2π−θ)=−cot⁡(θ)=cot⁡(−θ){\displaystyle \cot(2\pi -\theta )=-\cot(\theta )=\cot(-\theta )}

Shifts and periodicity

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Unit circle with a swept angle theta plotted at coordinates (a,b). As the swept angle is incremented by one-half pi (90 degrees), the coordinates are transformed to (-b,a). Another increment of one-half pi (180 degrees total) transforms the coordinates to (-a,-b). A final increment of one-half pi (270 degrees total) transforms the coordinates to (b,a).
Transformation of coordinates (a,b) when shifting the angleθ{\displaystyle \theta } in increments ofπ2{\displaystyle {\frac {\pi }{2}}}
Shift by one quarter periodShift by one half periodShift by full periods[4]Period
sin⁡(θ±π2)=±cos⁡θ{\displaystyle \sin(\theta \pm {\tfrac {\pi }{2}})=\pm \cos \theta }sin⁡(θ+π)=−sin⁡θ{\displaystyle \sin(\theta +\pi )=-\sin \theta }sin⁡(θ+k⋅2π)=+sin⁡θ{\displaystyle \sin(\theta +k\cdot 2\pi )=+\sin \theta }2π{\displaystyle 2\pi }
cos⁡(θ±π2)=∓sin⁡θ{\displaystyle \cos(\theta \pm {\tfrac {\pi }{2}})=\mp \sin \theta }cos⁡(θ+π)=−cos⁡θ{\displaystyle \cos(\theta +\pi )=-\cos \theta }cos⁡(θ+k⋅2π)=+cos⁡θ{\displaystyle \cos(\theta +k\cdot 2\pi )=+\cos \theta }2π{\displaystyle 2\pi }
csc⁡(θ±π2)=±sec⁡θ{\displaystyle \csc(\theta \pm {\tfrac {\pi }{2}})=\pm \sec \theta }csc⁡(θ+π)=−csc⁡θ{\displaystyle \csc(\theta +\pi )=-\csc \theta }csc⁡(θ+k⋅2π)=+csc⁡θ{\displaystyle \csc(\theta +k\cdot 2\pi )=+\csc \theta }2π{\displaystyle 2\pi }
sec⁡(θ±π2)=∓csc⁡θ{\displaystyle \sec(\theta \pm {\tfrac {\pi }{2}})=\mp \csc \theta }sec⁡(θ+π)=−sec⁡θ{\displaystyle \sec(\theta +\pi )=-\sec \theta }sec⁡(θ+k⋅2π)=+sec⁡θ{\displaystyle \sec(\theta +k\cdot 2\pi )=+\sec \theta }2π{\displaystyle 2\pi }
tan⁡(θ±π4)=tan⁡θ±11∓tan⁡θ{\displaystyle \tan(\theta \pm {\tfrac {\pi }{4}})={\tfrac {\tan \theta \pm 1}{1\mp \tan \theta }}}tan⁡(θ+π2)=−cot⁡θ{\displaystyle \tan(\theta +{\tfrac {\pi }{2}})=-\cot \theta }tan⁡(θ+k⋅π)=+tan⁡θ{\displaystyle \tan(\theta +k\cdot \pi )=+\tan \theta }π{\displaystyle \pi }
cot⁡(θ±π4)=cot⁡θ∓11±cot⁡θ{\displaystyle \cot(\theta \pm {\tfrac {\pi }{4}})={\tfrac {\cot \theta \mp 1}{1\pm \cot \theta }}}cot⁡(θ+π2)=−tan⁡θ{\displaystyle \cot(\theta +{\tfrac {\pi }{2}})=-\tan \theta }cot⁡(θ+k⋅π)=+cot⁡θ{\displaystyle \cot(\theta +k\cdot \pi )=+\cot \theta }π{\displaystyle \pi }

Signs

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The sign of trigonometric functions depends on quadrant of the angle. If−π<θ≤π{\displaystyle {-\pi }<\theta \leq \pi } andsgn is thesign function,

sgn⁡(sin⁡θ)=sgn⁡(csc⁡θ)={+1if  0<θ<π−1if  −π<θ<00if  θ∈{0,π}sgn⁡(cos⁡θ)=sgn⁡(sec⁡θ)={+1if  −12π<θ<12π−1if  −π<θ<−12π  or  12π<θ<π0if  θ∈{−12π,12π}sgn⁡(tan⁡θ)=sgn⁡(cot⁡θ)={+1if  −π<θ<−12π  or  0<θ<12π−1if  −12π<θ<0  or  12π<θ<π0if  θ∈{−12π,0,12π,π}{\displaystyle {\begin{aligned}\operatorname {sgn}(\sin \theta )=\operatorname {sgn}(\csc \theta )&={\begin{cases}+1&{\text{if}}\ \ 0<\theta <\pi \\-1&{\text{if}}\ \ {-\pi }<\theta <0\\0&{\text{if}}\ \ \theta \in \{0,\pi \}\end{cases}}\\[5mu]\operatorname {sgn}(\cos \theta )=\operatorname {sgn}(\sec \theta )&={\begin{cases}+1&{\text{if}}\ \ {-{\tfrac {1}{2}}\pi }<\theta <{\tfrac {1}{2}}\pi \\-1&{\text{if}}\ \ {-\pi }<\theta <-{\tfrac {1}{2}}\pi \ \ {\text{or}}\ \ {\tfrac {1}{2}}\pi <\theta <\pi \\0&{\text{if}}\ \ \theta \in {\bigl \{}{-{\tfrac {1}{2}}\pi },{\tfrac {1}{2}}\pi {\bigr \}}\end{cases}}\\[5mu]\operatorname {sgn}(\tan \theta )=\operatorname {sgn}(\cot \theta )&={\begin{cases}+1&{\text{if}}\ \ {-\pi }<\theta <-{\tfrac {1}{2}}\pi \ \ {\text{or}}\ \ 0<\theta <{\tfrac {1}{2}}\pi \\-1&{\text{if}}\ \ {-{\tfrac {1}{2}}\pi }<\theta <0\ \ {\text{or}}\ \ {\tfrac {1}{2}}\pi <\theta <\pi \\0&{\text{if}}\ \ \theta \in {\bigl \{}{-{\tfrac {1}{2}}\pi },0,{\tfrac {1}{2}}\pi ,\pi {\bigr \}}\end{cases}}\end{aligned}}}

The trigonometric functions are periodic with common period2π,{\displaystyle 2\pi ,} so for values ofθ outside the interval(−π,π],{\displaystyle ({-\pi },\pi ],} they take repeating values (see§ Shifts and periodicity above).

Angle sum and difference identities

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See also:Proofs of trigonometric identities § Angle sum identities, andSmall-angle approximation § Angle sum and difference
Geometric construction to derive angle sum trigonometric identities
Diagram showing the angle difference identities forsin⁡(α−β){\displaystyle \sin(\alpha -\beta )} andcos⁡(α−β){\displaystyle \cos(\alpha -\beta )}

These are also known as theangle addition and subtraction theorems (orformulae).sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡βsin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡βcos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡βcos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β{\displaystyle {\begin{aligned}\sin(\alpha +\beta )&=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\\sin(\alpha -\beta )&=\sin \alpha \cos \beta -\cos \alpha \sin \beta \\\cos(\alpha +\beta )&=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\\cos(\alpha -\beta )&=\cos \alpha \cos \beta +\sin \alpha \sin \beta \end{aligned}}}

The angle difference identities forsin⁡(α−β){\displaystyle \sin(\alpha -\beta )} andcos⁡(α−β){\displaystyle \cos(\alpha -\beta )} can be derived from the angle sum versions by substituting−β{\displaystyle -\beta } forβ{\displaystyle \beta } and using the facts thatsin⁡(−β)=−sin⁡(β){\displaystyle \sin(-\beta )=-\sin(\beta )} andcos⁡(−β)=cos⁡(β){\displaystyle \cos(-\beta )=\cos(\beta )}. They can also be derived by using a slightly modified version of the figure for the angle sum identities, both of which are shown here.

These identities are summarized in the first two rows of the following table, which also includes sum and difference identities for the other trigonometric functions.

Sinesin⁡(α±β){\displaystyle \sin(\alpha \pm \beta )}={\displaystyle =}sin⁡αcos⁡β±cos⁡αsin⁡β{\displaystyle \sin \alpha \cos \beta \pm \cos \alpha \sin \beta }[5][6]
Cosinecos⁡(α±β){\displaystyle \cos(\alpha \pm \beta )}={\displaystyle =}cos⁡αcos⁡β∓sin⁡αsin⁡β{\displaystyle \cos \alpha \cos \beta \mp \sin \alpha \sin \beta }[6][7]
Tangenttan⁡(α±β){\displaystyle \tan(\alpha \pm \beta )}={\displaystyle =}tan⁡α±tan⁡β1∓tan⁡αtan⁡β{\displaystyle {\frac {\tan \alpha \pm \tan \beta }{1\mp \tan \alpha \tan \beta }}}[6][8]
Cosecantcsc⁡(α±β){\displaystyle \csc(\alpha \pm \beta )}={\displaystyle =}sec⁡αsec⁡βcsc⁡αcsc⁡βsec⁡αcsc⁡β±csc⁡αsec⁡β{\displaystyle {\frac {\sec \alpha \sec \beta \csc \alpha \csc \beta }{\sec \alpha \csc \beta \pm \csc \alpha \sec \beta }}}[9]
Secantsec⁡(α±β){\displaystyle \sec(\alpha \pm \beta )}={\displaystyle =}sec⁡αsec⁡βcsc⁡αcsc⁡βcsc⁡αcsc⁡β∓sec⁡αsec⁡β{\displaystyle {\frac {\sec \alpha \sec \beta \csc \alpha \csc \beta }{\csc \alpha \csc \beta \mp \sec \alpha \sec \beta }}}[9]
Cotangentcot⁡(α±β){\displaystyle \cot(\alpha \pm \beta )}={\displaystyle =}cot⁡αcot⁡β∓1cot⁡β±cot⁡α{\displaystyle {\frac {\cot \alpha \cot \beta \mp 1}{\cot \beta \pm \cot \alpha }}}[6][10]
Arcsinearcsin⁡x±arcsin⁡y{\displaystyle \arcsin x\pm \arcsin y}={\displaystyle =}arcsin⁡(x1−y2±y1−x2y){\displaystyle \arcsin \left(x{\sqrt {1-y^{2}}}\pm y{\sqrt {1-x^{2}{\vphantom {y}}}}\right)}[11]
Arccosinearccos⁡x±arccos⁡y{\displaystyle \arccos x\pm \arccos y}={\displaystyle =}arccos⁡(xy∓(1−x2)(1−y2)){\displaystyle \arccos \left(xy\mp {\sqrt {\left(1-x^{2}\right)\left(1-y^{2}\right)}}\right)}[12]
Arctangentarctan⁡x±arctan⁡y{\displaystyle \arctan x\pm \arctan y}={\displaystyle =}arctan⁡(x±y1∓xy){\displaystyle \arctan \left({\frac {x\pm y}{1\mp xy}}\right)}[13]
Arccotangentarccot⁡x±arccot⁡y{\displaystyle \operatorname {arccot} x\pm \operatorname {arccot} y}={\displaystyle =}arccot⁡(xy∓1y±x){\displaystyle \operatorname {arccot} \left({\frac {xy\mp 1}{y\pm x}}\right)}

Sines and cosines of sums of infinitely many angles

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When the series∑i=1∞θi{\textstyle \sum _{i=1}^{\infty }\theta _{i}}converges absolutely then

sin(∑i=1∞θi)=∑odd k≥1(−1)k−12∑A⊆{1,2,3,…}|A|=k(∏i∈Asin⁡θi∏i∉Acos⁡θi)cos(∑i=1∞θi)=∑even k≥0(−1)k2∑A⊆{1,2,3,…}|A|=k(∏i∈Asin⁡θi∏i∉Acos⁡θi).{\displaystyle {\begin{aligned}{\sin }{\biggl (}\sum _{i=1}^{\infty }\theta _{i}{\biggl )}&=\sum _{{\text{odd}}\ k\geq 1}(-1)^{\frac {k-1}{2}}\!\!\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}{\biggl (}\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}{\biggr )}\\{\cos }{\biggl (}\sum _{i=1}^{\infty }\theta _{i}{\biggr )}&=\sum _{{\text{even}}\ k\geq 0}(-1)^{\frac {k}{2}}\,\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}{\biggl (}\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}{\biggr )}.\end{aligned}}}

Because the series∑i=1∞θi{\textstyle \sum _{i=1}^{\infty }\theta _{i}} converges absolutely, it is necessarily the case thatlimi→∞θi=0,{\textstyle \lim _{i\to \infty }\theta _{i}=0,}limi→∞sin⁡θi=0,{\textstyle \lim _{i\to \infty }\sin \theta _{i}=0,} andlimi→∞cos⁡θi=1.{\textstyle \lim _{i\to \infty }\cos \theta _{i}=1.} Particularly, in these two identities, an asymmetry appears that is not seen in the case of sums of finitely many angles: in each product, there are only finitely many sine factors but there arecofinitely many cosine factors. Terms with infinitely many sine factors would necessarily be equal to zero.

When only finitely many of the anglesθi{\displaystyle \theta _{i}} are nonzero then only finitely many of the terms on the right side are nonzero because all but finitely many sine factors vanish. Furthermore, in each term all but finitely many of the cosine factors are unity.

Tangents and cotangents of sums

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Letek{\displaystyle e_{k}} (fork=0,1,2,3,…{\displaystyle k=0,1,2,3,\ldots }) be thekth-degreeelementary symmetric polynomial in the variablesxi=tan⁡θi{\displaystyle x_{i}=\tan \theta _{i}}fori=0,1,2,3,…,{\displaystyle i=0,1,2,3,\ldots ,} that is,

e0=1e1=∑ixi=∑itan⁡θie2=∑i<jxixj=∑i<jtan⁡θitan⁡θje3=∑i<j<kxixjxk=∑i<j<ktan⁡θitan⁡θjtan⁡θk  ⋮  ⋮{\displaystyle {\begin{aligned}e_{0}&=1\\[6pt]e_{1}&=\sum _{i}x_{i}&&=\sum _{i}\tan \theta _{i}\\[6pt]e_{2}&=\sum _{i<j}x_{i}x_{j}&&=\sum _{i<j}\tan \theta _{i}\tan \theta _{j}\\[6pt]e_{3}&=\sum _{i<j<k}x_{i}x_{j}x_{k}&&=\sum _{i<j<k}\tan \theta _{i}\tan \theta _{j}\tan \theta _{k}\\&\ \ \vdots &&\ \ \vdots \end{aligned}}}

Then

tan⁡(∑iθi)=e1−e3+e5−⋯e0−e2+e4−⋯.{\displaystyle \tan {\Bigl (}\sum _{i}\theta _{i}{\Bigr )}={\frac {e_{1}-e_{3}+e_{5}-\cdots }{e_{0}-e_{2}+e_{4}-\cdots }}.}This can be shown by using the sine and cosine sum formulae above:tan⁡(∑iθi)=sin(∑iθi)/∏icos⁡θicos(∑iθi)/∏icos⁡θi=∑odd k≥1(−1)k−12∑A⊆{1,2,3,…}|A|=k∏i∈Atan⁡θi∑even k≥0 (−1)k2  ∑A⊆{1,2,3,…}|A|=k∏i∈Atan⁡θi=e1−e3+e5−⋯e0−e2+e4−⋯cot⁡(∑iθi)=e0−e2+e4−⋯e1−e3+e5−⋯{\displaystyle {\begin{aligned}\tan {\Bigl (}\sum _{i}\theta _{i}{\Bigr )}&={\frac {{\sin }{\bigl (}\sum _{i}\theta _{i}{\bigr )}/\prod _{i}\cos \theta _{i}}{{\cos }{\bigl (}\sum _{i}\theta _{i}{\bigr )}/\prod _{i}\cos \theta _{i}}}\\[10pt]&={\frac {\displaystyle \sum _{{\text{odd}}\ k\geq 1}(-1)^{\frac {k-1}{2}}\sum _{\begin{smallmatrix}A\subseteq \{1,2,3,\dots \}\\\left|A\right|=k\end{smallmatrix}}\prod _{i\in A}\tan \theta _{i}}{\displaystyle \sum _{{\text{even}}\ k\geq 0}~(-1)^{\frac {k}{2}}~~\sum _{\begin{smallmatrix}A\subseteq \{1,2,3,\dots \}\\\left|A\right|=k\end{smallmatrix}}\prod _{i\in A}\tan \theta _{i}}}={\frac {e_{1}-e_{3}+e_{5}-\cdots }{e_{0}-e_{2}+e_{4}-\cdots }}\\[10pt]\cot {\Bigl (}\sum _{i}\theta _{i}{\Bigr )}&={\frac {e_{0}-e_{2}+e_{4}-\cdots }{e_{1}-e_{3}+e_{5}-\cdots }}\end{aligned}}}

The number of terms on the right side depends on the number of terms on the left side.

For example:tan⁡(θ1+θ2)=e1e0−e2=x1+x21 − x1x2=tan⁡θ1+tan⁡θ21 − tan⁡θ1tan⁡θ2,tan⁡(θ1+θ2+θ3)=e1−e3e0−e2=(x1+x2+x3) − (x1x2x3)1 − (x1x2+x1x3+x2x3),tan⁡(θ1+θ2+θ3+θ4)=e1−e3e0−e2+e4=(x1+x2+x3+x4) − (x1x2x3+x1x2x4+x1x3x4+x2x3x4)1 − (x1x2+x1x3+x1x4+x2x3+x2x4+x3x4) + (x1x2x3x4),{\displaystyle {\begin{aligned}\tan(\theta _{1}+\theta _{2})&={\frac {e_{1}}{e_{0}-e_{2}}}={\frac {x_{1}+x_{2}}{1\ -\ x_{1}x_{2}}}={\frac {\tan \theta _{1}+\tan \theta _{2}}{1\ -\ \tan \theta _{1}\tan \theta _{2}}},\\[8pt]\tan(\theta _{1}+\theta _{2}+\theta _{3})&={\frac {e_{1}-e_{3}}{e_{0}-e_{2}}}={\frac {(x_{1}+x_{2}+x_{3})\ -\ (x_{1}x_{2}x_{3})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3})}},\\[8pt]\tan(\theta _{1}+\theta _{2}+\theta _{3}+\theta _{4})&={\frac {e_{1}-e_{3}}{e_{0}-e_{2}+e_{4}}}\\[8pt]&={\frac {(x_{1}+x_{2}+x_{3}+x_{4})\ -\ (x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4})\ +\ (x_{1}x_{2}x_{3}x_{4})}},\end{aligned}}}

and so on. The case of only finitely many terms can be proved bymathematical induction.[14] The case of infinitely many terms can be proved by using some elementary inequalities.[15]

Linear fractional transformations of tangents, related to tangents of sums

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Supposea,b,c,d,p,q∈R{\textstyle a,b,c,d,p,q\in \mathbb {R} } andi=−1{\textstyle i={\sqrt {-1}}} and

ai+bci+d=pi+q{\displaystyle {\frac {ai+b}{ci+d}}=pi+q}

and letφ{\textstyle \varphi } be any number for whichtan⁡φ=c/d.{\textstyle \tan \varphi =c/d.}Suppose thata/c≠b/d{\textstyle a/c\neq b/d} so that the forgoing fraction cannot be0/0{\textstyle 0/0}. Then for allθ∈R{\textstyle \theta \in \mathbb {R} }[16]

atan⁡θ+bctan⁡θ+d=ptan⁡(θ−φ)+q.{\displaystyle {\frac {a\tan \theta +b}{c\tan \theta +d}}=p\tan(\theta -\varphi )+q.}

(In case the denominator of this fraction is 0, we take the value of the fraction to be∞{\textstyle \infty }, where the symbol∞{\textstyle \infty } does not mean either+∞{\textstyle +\infty } or−∞{\textstyle -\infty }, but is the∞{\textstyle \infty } that is approached by going in either the positive or the negative direction, making the completion of the lineR∪{∞}{\textstyle \mathbb {R} \cup \{\,\infty \,\}} topologically a circle.)

From this identity it can be shown to follow quickly that the family of allCauchy-distributed random variables is closed under linear fractional transformations, a result known since 1976.[17]

Secants and cosecants of sums

[edit]

sec(∑iθi)=∏isec⁡θie0−e2+e4−⋯csc(∑iθi)=∏isec⁡θie1−e3+e5−⋯{\displaystyle {\begin{aligned}{\sec }{\Bigl (}\sum _{i}\theta _{i}{\Bigr )}&={\frac {\prod _{i}\sec \theta _{i}}{e_{0}-e_{2}+e_{4}-\cdots }}\\[8pt]{\csc }{\Bigl (}\sum _{i}\theta _{i}{\Bigr )}&={\frac {\prod _{i}\sec \theta _{i}}{e_{1}-e_{3}+e_{5}-\cdots }}\end{aligned}}}

whereek{\displaystyle e_{k}} is thekth-degreeelementary symmetric polynomial in then variablesxi=tan⁡θi,{\displaystyle x_{i}=\tan \theta _{i},}i=1,…,n,{\displaystyle i=1,\ldots ,n,} and the number of terms in the denominator and the number of factors in the product in the numerator depend on the number of terms in the sum on the left.[18] The case of only finitely many terms can be proved by mathematical induction on the number of such terms.

For example,

sec⁡(α+β+γ)=sec⁡αsec⁡βsec⁡γ1−tan⁡αtan⁡β−tan⁡αtan⁡γ−tan⁡βtan⁡γcsc⁡(α+β+γ)=sec⁡αsec⁡βsec⁡γtan⁡α+tan⁡β+tan⁡γ−tan⁡αtan⁡βtan⁡γ.{\displaystyle {\begin{aligned}\sec(\alpha +\beta +\gamma )&={\frac {\sec \alpha \sec \beta \sec \gamma }{1-\tan \alpha \tan \beta -\tan \alpha \tan \gamma -\tan \beta \tan \gamma }}\\[8pt]\csc(\alpha +\beta +\gamma )&={\frac {\sec \alpha \sec \beta \sec \gamma }{\tan \alpha +\tan \beta +\tan \gamma -\tan \alpha \tan \beta \tan \gamma }}.\end{aligned}}}

Ptolemy's theorem

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Main article:Ptolemy's theorem
See also:History of trigonometry § Classical antiquity
Diagram illustrating the relation between Ptolemy's theorem and the angle sum trig identity for sine. Ptolemy's theorem states that the sum of the products of the lengths of opposite sides is equal to the product of the lengths of the diagonals. When those side-lengths are expressed in terms of the sin and cos values shown in the figure above, this yields the angle sum trigonometric identity for sine:sin(α +β) = sinα cosβ + cosα sinβ.

Ptolemy's theorem is important in the history of trigonometric identities, as it is how results equivalent to the sum and difference formulas for sine and cosine were first proved. It states that in acyclic quadrilateralABCD{\displaystyle ABCD}, as shown in the accompanying figure, the sum of the products of the lengths of opposite sides is equal to the product of the lengths of the diagonals. In the special cases of one of the diagonals or sides being a diameter of the circle, this theorem gives rise directly to the angle sum and difference trigonometric identities.[19] The relationship follows most easily when the circle is constructed to have a diameter of length one, as shown here.

ByThales's theorem,∠DAB{\displaystyle \angle DAB} and∠DCB{\displaystyle \angle DCB} are both right angles. The right-angled trianglesDAB{\displaystyle DAB} andDCB{\displaystyle DCB} both share the hypotenuseBD¯{\displaystyle {\overline {BD}}} of length 1. Thus, the sideAB¯=sin⁡α{\displaystyle {\overline {AB}}=\sin \alpha },AD¯=cos⁡α{\displaystyle {\overline {AD}}=\cos \alpha },BC¯=sin⁡β{\displaystyle {\overline {BC}}=\sin \beta } andCD¯=cos⁡β{\displaystyle {\overline {CD}}=\cos \beta }.

By theinscribed angle theorem, thecentral angle subtended by the chordAC¯{\displaystyle {\overline {AC}}} at the circle's center is twice the angle∠ADC{\displaystyle \angle ADC}, i.e.2(α+β){\displaystyle 2(\alpha +\beta )}. Therefore, the symmetrical pair of red triangles each has the angleα+β{\displaystyle \alpha +\beta } at the center. Each of these triangles has ahypotenuse of length12{\textstyle {\frac {1}{2}}}, so the length ofAC¯{\displaystyle {\overline {AC}}} is2×12sin⁡(α+β){\textstyle 2\times {\frac {1}{2}}\sin(\alpha +\beta )}, i.e. simplysin⁡(α+β){\displaystyle \sin(\alpha +\beta )}. The quadrilateral's other diagonal is the diameter of length 1, so the product of the diagonals' lengths is alsosin⁡(α+β){\displaystyle \sin(\alpha +\beta )}.

When these values are substituted into the statement of Ptolemy's theorem that|AC¯|⋅|BD¯|=|AB¯|⋅|CD¯|+|AD¯|⋅|BC¯|{\displaystyle |{\overline {AC}}|\cdot |{\overline {BD}}|=|{\overline {AB}}|\cdot |{\overline {CD}}|+|{\overline {AD}}|\cdot |{\overline {BC}}|}, this yields the angle sum trigonometric identity for sine:sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β{\displaystyle \sin(\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta }. The angle difference formula forsin⁡(α−β){\displaystyle \sin(\alpha -\beta )} can be similarly derived by letting the sideCD¯{\displaystyle {\overline {CD}}} serve as a diameter instead ofBD¯{\displaystyle {\overline {BD}}}.[19]

Multiple-angle and half-angle formulae

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Tn is thenthChebyshev polynomialcos⁡(nθ)=Tn(cos⁡θ){\displaystyle \cos(n\theta )=T_{n}(\cos \theta )}[20]
de Moivre's formula,i is theimaginary unitcos⁡(nθ)+isin⁡(nθ)=(cos⁡θ+isin⁡θ)n{\displaystyle \cos(n\theta )+i\sin(n\theta )=(\cos \theta +i\sin \theta )^{n}}[21]

Multiple-angle formulae

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Double-angle formulae

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Visual demonstration of the double-angle formula for sine. For the above isosceles triangle with unit sides and angle2θ{\displaystyle 2\theta }, the area⁠1/2⁠ × base × height is calculated in two orientations. When upright, the area issin⁡θcos⁡θ{\displaystyle \sin \theta \cos \theta }. When on its side, the same area is12sin⁡2θ{\textstyle {\frac {1}{2}}\sin 2\theta }. Therefore,sin⁡2θ=2sin⁡θcos⁡θ.{\displaystyle \sin 2\theta =2\sin \theta \cos \theta .}

Formulae for twice an angle.[22]

Triple-angle formulae

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Formulae for triple angles.[22]

Multiple-angle formulae

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Formulae for multiple angles.[23]

Chebyshev method

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TheChebyshev method is arecursivealgorithm for finding thenth multiple angle formula knowing the(n−1){\displaystyle (n-1)}th and(n−2){\displaystyle (n-2)}th values.[24]

cos⁡(nx){\displaystyle \cos(nx)} can be computed fromcos⁡((n−1)x){\displaystyle \cos((n-1)x)},cos⁡((n−2)x){\displaystyle \cos((n-2)x)}, andcos⁡(x){\displaystyle \cos(x)} with

cos⁡(nx)=2cos⁡xcos⁡((n−1)x)−cos⁡((n−2)x).{\displaystyle \cos(nx)=2\cos x\cos((n-1)x)-\cos((n-2)x).}

This can be proved by adding together the formulae

cos⁡((n−1)x+x)=cos⁡((n−1)x)cos⁡x−sin⁡((n−1)x)sin⁡xcos⁡((n−1)x−x)=cos⁡((n−1)x)cos⁡x+sin⁡((n−1)x)sin⁡x{\displaystyle {\begin{aligned}\cos((n-1)x+x)&=\cos((n-1)x)\cos x-\sin((n-1)x)\sin x\\\cos((n-1)x-x)&=\cos((n-1)x)\cos x+\sin((n-1)x)\sin x\end{aligned}}}

It follows by induction thatcos⁡(nx){\displaystyle \cos(nx)} is a polynomial ofcos⁡x,{\displaystyle \cos x,} the so-called Chebyshev polynomial of the first kind, seeChebyshev polynomials#Trigonometric definition.

Similarly,sin⁡(nx){\displaystyle \sin(nx)} can be computed fromsin⁡((n−1)x),{\displaystyle \sin((n-1)x),}sin⁡((n−2)x),{\displaystyle \sin((n-2)x),} andcos⁡x{\displaystyle \cos x} withsin⁡(nx)=2cos⁡xsin⁡((n−1)x)−sin⁡((n−2)x){\displaystyle \sin(nx)=2\cos x\sin((n-1)x)-\sin((n-2)x)}This can be proved by adding formulae forsin⁡((n−1)x+x){\displaystyle \sin((n-1)x+x)} andsin⁡((n−1)x−x).{\displaystyle \sin((n-1)x-x).}

Serving a purpose similar to that of the Chebyshev method, for the tangent we can write:

tan⁡(nx)=tan⁡((n−1)x)+tan⁡x1−tan⁡((n−1)x)tan⁡x.{\displaystyle \tan(nx)={\frac {\tan((n-1)x)+\tan x}{1-\tan((n-1)x)\tan x}}\,.}

Half-angle formulae

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sin⁡θ2=sgn⁡(sin⁡θ2)1−cos⁡θ2cos⁡θ2=sgn⁡(cos⁡θ2)1+cos⁡θ2tan⁡θ2=1−cos⁡θsin⁡θ=sin⁡θ1+cos⁡θ=csc⁡θ−cot⁡θ=tan⁡θ1+sec⁡θ=sgn⁡(sin⁡θ)1−cos⁡θ1+cos⁡θ=−1+sgn⁡(cos⁡θ)1+tan2⁡θtan⁡θcot⁡θ2=1+cos⁡θsin⁡θ=sin⁡θ1−cos⁡θ=csc⁡θ+cot⁡θ=sgn⁡(sin⁡θ)1+cos⁡θ1−cos⁡θsec⁡θ2=sgn⁡(cos⁡θ2)21+cos⁡θcsc⁡θ2=sgn⁡(sin⁡θ2)21−cos⁡θ{\displaystyle {\begin{aligned}\sin {\frac {\theta }{2}}&=\operatorname {sgn} \left(\sin {\frac {\theta }{2}}\right){\sqrt {\frac {1-\cos \theta }{2}}}\\[3pt]\cos {\frac {\theta }{2}}&=\operatorname {sgn} \left(\cos {\frac {\theta }{2}}\right){\sqrt {\frac {1+\cos \theta }{2}}}\\[3pt]\tan {\frac {\theta }{2}}&={\frac {1-\cos \theta }{\sin \theta }}={\frac {\sin \theta }{1+\cos \theta }}=\csc \theta -\cot \theta ={\frac {\tan \theta }{1+\sec {\theta }}}\\[6mu]&=\operatorname {sgn}(\sin \theta ){\sqrt {\frac {1-\cos \theta }{1+\cos \theta }}}={\frac {-1+\operatorname {sgn}(\cos \theta ){\sqrt {1+\tan ^{2}\theta }}}{\tan \theta }}\\[3pt]\cot {\frac {\theta }{2}}&={\frac {1+\cos \theta }{\sin \theta }}={\frac {\sin \theta }{1-\cos \theta }}=\csc \theta +\cot \theta =\operatorname {sgn}(\sin \theta ){\sqrt {\frac {1+\cos \theta }{1-\cos \theta }}}\\\sec {\frac {\theta }{2}}&=\operatorname {sgn} \left(\cos {\frac {\theta }{2}}\right){\sqrt {\frac {2}{1+\cos \theta }}}\\\csc {\frac {\theta }{2}}&=\operatorname {sgn} \left(\sin {\frac {\theta }{2}}\right){\sqrt {\frac {2}{1-\cos \theta }}}\\\end{aligned}}}[25][26]

Alsotan⁡η±θ2=sin⁡η±sin⁡θcos⁡η+cos⁡θtan⁡(θ2+π4)=sec⁡θ+tan⁡θ1−sin⁡θ1+sin⁡θ=|1−tan⁡θ2||1+tan⁡θ2|{\displaystyle {\begin{aligned}\tan {\frac {\eta \pm \theta }{2}}&={\frac {\sin \eta \pm \sin \theta }{\cos \eta +\cos \theta }}\\[3pt]\tan \left({\frac {\theta }{2}}+{\frac {\pi }{4}}\right)&=\sec \theta +\tan \theta \\[3pt]{\sqrt {\frac {1-\sin \theta }{1+\sin \theta }}}&={\frac {\left|1-\tan {\frac {\theta }{2}}\right|}{\left|1+\tan {\frac {\theta }{2}}\right|}}\end{aligned}}}

Table

[edit]
See also:Tangent half-angle formula

These can be shown by using either the sum and difference identities or the multiple-angle formulae.

SineCosineTangentCotangent
Double-angle formula[27][28]sin⁡(2θ)=2sin⁡θcos⁡θ =2tan⁡θ1+tan2⁡θ{\displaystyle {\begin{aligned}\sin(2\theta )&=2\sin \theta \cos \theta \ \\&={\frac {2\tan \theta }{1+\tan ^{2}\theta }}\end{aligned}}}cos⁡(2θ)=cos2⁡θ−sin2⁡θ=2cos2⁡θ−1=1−2sin2⁡θ=1−tan2⁡θ1+tan2⁡θ{\displaystyle {\begin{aligned}\cos(2\theta )&=\cos ^{2}\theta -\sin ^{2}\theta \\&=2\cos ^{2}\theta -1\\&=1-2\sin ^{2}\theta \\&={\frac {1-\tan ^{2}\theta }{1+\tan ^{2}\theta }}\end{aligned}}}tan⁡(2θ)=2tan⁡θ1−tan2⁡θ{\displaystyle \tan(2\theta )={\frac {2\tan \theta }{1-\tan ^{2}\theta }}}cot⁡(2θ)=cot2⁡θ−12cot⁡θ{\displaystyle \cot(2\theta )={\frac {\cot ^{2}\theta -1}{2\cot \theta }}}
Triple-angle formula[20][29]sin⁡(3θ)=−sin3⁡θ+3cos2⁡θsin⁡θ=−4sin3⁡θ+3sin⁡θ{\displaystyle {\begin{aligned}\sin(3\theta )&=-\sin ^{3}\theta +3\cos ^{2}\theta \sin \theta \\&=-4\sin ^{3}\theta +3\sin \theta \end{aligned}}}cos⁡(3θ)=cos3⁡θ−3sin2⁡θcos⁡θ=4cos3⁡θ−3cos⁡θ{\displaystyle {\begin{aligned}\cos(3\theta )&=\cos ^{3}\theta -3\sin ^{2}\theta \cos \theta \\&=4\cos ^{3}\theta -3\cos \theta \end{aligned}}}tan⁡(3θ)=3tan⁡θ−tan3⁡θ1−3tan2⁡θ{\displaystyle \tan(3\theta )={\frac {3\tan \theta -\tan ^{3}\theta }{1-3\tan ^{2}\theta }}}cot⁡(3θ)=3cot⁡θ−cot3⁡θ1−3cot2⁡θ{\displaystyle \cot(3\theta )={\frac {3\cot \theta -\cot ^{3}\theta }{1-3\cot ^{2}\theta }}}
Half-angle formula[25][26]sin⁡θ2=sgn⁡(sin⁡θ2)1−cos⁡θ2(or sin2⁡θ2=1−cos⁡θ2){\displaystyle {\begin{aligned}&\sin {\frac {\theta }{2}}=\operatorname {sgn} \left(\sin {\frac {\theta }{2}}\right){\sqrt {\frac {1-\cos \theta }{2}}}\\\\&\left({\text{or }}\sin ^{2}{\frac {\theta }{2}}={\frac {1-\cos \theta }{2}}\right)\end{aligned}}}cos⁡θ2=sgn⁡(cos⁡θ2)1+cos⁡θ2(or cos2⁡θ2=1+cos⁡θ2){\displaystyle {\begin{aligned}&\cos {\frac {\theta }{2}}=\operatorname {sgn} \left(\cos {\frac {\theta }{2}}\right){\sqrt {\frac {1+\cos \theta }{2}}}\\\\&\left({\text{or }}\cos ^{2}{\frac {\theta }{2}}={\frac {1+\cos \theta }{2}}\right)\end{aligned}}}tan⁡θ2=csc⁡θ−cot⁡θ=±1−cos⁡θ1+cos⁡θ=sin⁡θ1+cos⁡θ=1−cos⁡θsin⁡θtan⁡η+θ2=sin⁡η+sin⁡θcos⁡η+cos⁡θtan⁡(θ2+π4)=sec⁡θ+tan⁡θ1−sin⁡θ1+sin⁡θ=|1−tan⁡θ2||1+tan⁡θ2|tan⁡θ2=tan⁡θ1+1+tan2⁡θfor θ∈(−π2,π2){\displaystyle {\begin{aligned}\tan {\frac {\theta }{2}}&=\csc \theta -\cot \theta \\&=\pm \,{\sqrt {\frac {1-\cos \theta }{1+\cos \theta }}}\\[3pt]&={\frac {\sin \theta }{1+\cos \theta }}\\[3pt]&={\frac {1-\cos \theta }{\sin \theta }}\\[5pt]\tan {\frac {\eta +\theta }{2}}&={\frac {\sin \eta +\sin \theta }{\cos \eta +\cos \theta }}\\[5pt]\tan \left({\frac {\theta }{2}}+{\frac {\pi }{4}}\right)&=\sec \theta +\tan \theta \\[5pt]{\sqrt {\frac {1-\sin \theta }{1+\sin \theta }}}&={\frac {\left|1-\tan {\frac {\theta }{2}}\right|}{\left|1+\tan {\frac {\theta }{2}}\right|}}\\[5pt]\tan {\frac {\theta }{2}}&={\frac {\tan \theta }{1+{\sqrt {1+\tan ^{2}\theta }}}}\\&{\text{for }}\theta \in \left(-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}\right)\end{aligned}}}cot⁡θ2=csc⁡θ+cot⁡θ=±1+cos⁡θ1−cos⁡θ=sin⁡θ1−cos⁡θ=1+cos⁡θsin⁡θ{\displaystyle {\begin{aligned}\cot {\frac {\theta }{2}}&=\csc \theta +\cot \theta \\&=\pm \,{\sqrt {\frac {1+\cos \theta }{1-\cos \theta }}}\\[3pt]&={\frac {\sin \theta }{1-\cos \theta }}\\[4pt]&={\frac {1+\cos \theta }{\sin \theta }}\end{aligned}}}

The fact that the triple-angle formula for sine and cosine only involves powers of a single function allows one to relate the geometric problem of acompass and straightedge construction ofangle trisection to the algebraic problem of solving acubic equation, which allows one to prove thattrisection is in general impossible using the given tools.

A formula for computing the trigonometric identities for the one-third angle exists, but it requires finding the zeroes of thecubic equation4x3 − 3x +d = 0, wherex{\displaystyle x} is the value of the cosine function at the one-third angle andd is the known value of the cosine function at the full angle. However, thediscriminant of this equation is positive, so this equation has three real roots (of which only one is the solution for the cosine of the one-third angle).None of these solutions are reducible to a realalgebraic expression, as they use intermediate complex numbers under thecube roots.

Power-reduction formulae

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Obtained by solving the second and third versions of the cosine double-angle formula.

SineCosineOther
sin2⁡θ=1−cos⁡(2θ)2{\displaystyle \sin ^{2}\theta ={\frac {1-\cos(2\theta )}{2}}}cos2⁡θ=1+cos⁡(2θ)2{\displaystyle \cos ^{2}\theta ={\frac {1+\cos(2\theta )}{2}}}sin2⁡θcos2⁡θ=1−cos⁡(4θ)8{\displaystyle \sin ^{2}\theta \cos ^{2}\theta ={\frac {1-\cos(4\theta )}{8}}}
sin3⁡θ=3sin⁡θ−sin⁡(3θ)4{\displaystyle \sin ^{3}\theta ={\frac {3\sin \theta -\sin(3\theta )}{4}}}cos3⁡θ=3cos⁡θ+cos⁡(3θ)4{\displaystyle \cos ^{3}\theta ={\frac {3\cos \theta +\cos(3\theta )}{4}}}sin3⁡θcos3⁡θ=3sin⁡(2θ)−sin⁡(6θ)32{\displaystyle \sin ^{3}\theta \cos ^{3}\theta ={\frac {3\sin(2\theta )-\sin(6\theta )}{32}}}
sin4⁡θ=3−4cos⁡(2θ)+cos⁡(4θ)8{\displaystyle \sin ^{4}\theta ={\frac {3-4\cos(2\theta )+\cos(4\theta )}{8}}}cos4⁡θ=3+4cos⁡(2θ)+cos⁡(4θ)8{\displaystyle \cos ^{4}\theta ={\frac {3+4\cos(2\theta )+\cos(4\theta )}{8}}}sin4⁡θcos4⁡θ=3−4cos⁡(4θ)+cos⁡(8θ)128{\displaystyle \sin ^{4}\theta \cos ^{4}\theta ={\frac {3-4\cos(4\theta )+\cos(8\theta )}{128}}}
sin5⁡θ=10sin⁡θ−5sin⁡(3θ)+sin⁡(5θ)16{\displaystyle \sin ^{5}\theta ={\frac {10\sin \theta -5\sin(3\theta )+\sin(5\theta )}{16}}}cos5⁡θ=10cos⁡θ+5cos⁡(3θ)+cos⁡(5θ)16{\displaystyle \cos ^{5}\theta ={\frac {10\cos \theta +5\cos(3\theta )+\cos(5\theta )}{16}}}sin5⁡θcos5⁡θ=10sin⁡(2θ)−5sin⁡(6θ)+sin⁡(10θ)512{\displaystyle \sin ^{5}\theta \cos ^{5}\theta ={\frac {10\sin(2\theta )-5\sin(6\theta )+\sin(10\theta )}{512}}}
Cosine power-reduction formula: an illustrative diagram. The red, orange and blue triangles are all similar, and the red and orange triangles are congruent. The hypotenuseAD¯{\displaystyle {\overline {AD}}} of the blue triangle has length2cos⁡θ{\displaystyle 2\cos \theta }. The angle∠DAE{\displaystyle \angle DAE} isθ{\displaystyle \theta }, so the baseAE¯{\displaystyle {\overline {AE}}} of that triangle has length2cos2⁡θ{\displaystyle 2\cos ^{2}\theta }. That length is also equal to the summed lengths ofBD¯{\displaystyle {\overline {BD}}} andAF¯{\displaystyle {\overline {AF}}}, i.e.1+cos⁡(2θ){\displaystyle 1+\cos(2\theta )}. Therefore,2cos2⁡θ=1+cos⁡(2θ){\displaystyle 2\cos ^{2}\theta =1+\cos(2\theta )}. Dividing both sides by2{\displaystyle 2} yields the power-reduction formula for cosine:cos2⁡θ={\displaystyle \cos ^{2}\theta =}12(1+cos⁡(2θ)){\textstyle {\frac {1}{2}}(1+\cos(2\theta ))}. The half-angle formula for cosine can be obtained by replacingθ{\displaystyle \theta } withθ/2{\displaystyle \theta /2} and taking the square-root of both sides:cos⁡(θ/2)=±(1+cos⁡θ)/2.{\textstyle \cos \left(\theta /2\right)=\pm {\sqrt {\left(1+\cos \theta \right)/2}}.}
Sine power-reduction formula: an illustrative diagram. The shaded blue and green triangles, and the red-outlined triangleEBD{\displaystyle EBD} are all right-angled and similar, and all contain the angleθ{\displaystyle \theta }. The hypotenuseBD¯{\displaystyle {\overline {BD}}} of the red-outlined triangle has length2sin⁡θ{\displaystyle 2\sin \theta }, so its sideDE¯{\displaystyle {\overline {DE}}} has length2sin2⁡θ{\displaystyle 2\sin ^{2}\theta }. The line segmentAE¯{\displaystyle {\overline {AE}}} has lengthcos⁡2θ{\displaystyle \cos 2\theta } and sum of the lengths ofAE¯{\displaystyle {\overline {AE}}} andDE¯{\displaystyle {\overline {DE}}} equals the length ofAD¯{\displaystyle {\overline {AD}}}, which is 1. Therefore,cos⁡2θ+2sin2⁡θ=1{\displaystyle \cos 2\theta +2\sin ^{2}\theta =1}. Subtractingcos⁡2θ{\displaystyle \cos 2\theta } from both sides and dividing by 2 by two yields the power-reduction formula for sine:sin2⁡θ={\displaystyle \sin ^{2}\theta =}12(1−cos⁡(2θ)){\textstyle {\frac {1}{2}}(1-\cos(2\theta ))}. The half-angle formula for sine can be obtained by replacingθ{\displaystyle \theta } withθ/2{\displaystyle \theta /2} and taking the square-root of both sides:sin⁡(θ/2)=±(1−cos⁡θ)/2.{\textstyle \sin \left(\theta /2\right)=\pm {\sqrt {\left(1-\cos \theta \right)/2}}.} Note that this figure also illustrates, in the vertical line segmentEB¯{\displaystyle {\overline {EB}}}, thatsin⁡2θ=2sin⁡θcos⁡θ{\displaystyle \sin 2\theta =2\sin \theta \cos \theta }.

In general terms of powers ofsin⁡θ{\displaystyle \sin \theta } orcos⁡θ{\displaystyle \cos \theta } the following is true, and can be deduced usingDe Moivre's formula,Euler's formula and thebinomial theorem.

ifn is ...cosn⁡θ{\displaystyle \cos ^{n}\theta }sinn⁡θ{\displaystyle \sin ^{n}\theta }
n is oddcosn⁡θ=22n∑k=0n−12(nk)cos⁡((n−2k)θ){\displaystyle \cos ^{n}\theta ={\frac {2}{2^{n}}}\sum _{k=0}^{\frac {n-1}{2}}{\binom {n}{k}}\cos {{\big (}(n-2k)\theta {\big )}}}sinn⁡θ=22n∑k=0n−12(−1)(n−12−k)(nk)sin⁡((n−2k)θ){\displaystyle \sin ^{n}\theta ={\frac {2}{2^{n}}}\sum _{k=0}^{\frac {n-1}{2}}(-1)^{\left({\frac {n-1}{2}}-k\right)}{\binom {n}{k}}\sin {{\big (}(n-2k)\theta {\big )}}}
n is evencosn⁡θ=12n(nn2)+22n∑k=0n2−1(nk)cos⁡((n−2k)θ){\displaystyle \cos ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}{\binom {n}{k}}\cos {{\big (}(n-2k)\theta {\big )}}}sinn⁡θ=12n(nn2)+22n∑k=0n2−1(−1)(n2−k)(nk)cos⁡((n−2k)θ){\displaystyle \sin ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}(-1)^{\left({\frac {n}{2}}-k\right)}{\binom {n}{k}}\cos {{\big (}(n-2k)\theta {\big )}}}

Product-to-sum and sum-to-product identities

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Proof of the sum-and-difference-to-product cosine identity for prosthaphaeresis calculations using anisosceles triangle

The product-to-sum identities[30] orprosthaphaeresis formulae can be proven by expanding their right-hand sides using theangle addition theorems. Historically, the first four of these were known asWerner's formulas, afterJohannes Werner who used them for astronomical calculations.[31] Seeamplitude modulation for an application of the product-to-sum formulae, andbeat (acoustics) andphase detector for applications of the sum-to-product formulae.

Product-to-sum identities

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The product of two sines or cosines of different angles can be converted to a sum of trigonometric functions of a sum and difference of those angles:

cos⁡θcos⁡φ=12( cos⁡(θ−φ)+cos⁡(θ+φ)),sin⁡θsin⁡φ=12( cos⁡(θ−φ)−cos⁡(θ+φ)),sin⁡θcos⁡φ=12( sin⁡(θ+φ)+sin⁡(θ−φ)),cos⁡θsin⁡φ=12( sin⁡(θ+φ)−sin⁡(θ−φ)).{\displaystyle {\begin{aligned}\cos \theta \,\cos \varphi &={\tfrac {1}{2}}{\bigl (}\!\!~\cos(\theta -\varphi )+\cos(\theta +\varphi ){\bigr )},\\[5mu]\sin \theta \,\sin \varphi &={\tfrac {1}{2}}{\bigl (}\!\!~\cos(\theta -\varphi )-\cos(\theta +\varphi ){\bigr )},\\[5mu]\sin \theta \,\cos \varphi &={\tfrac {1}{2}}{\bigl (}\!\!~\sin(\theta +\varphi )+\sin(\theta -\varphi ){\bigr )},\\[5mu]\cos \theta \,\sin \varphi &={\tfrac {1}{2}}{\bigl (}\!\!~\sin(\theta +\varphi )-\sin(\theta -\varphi ){\bigr )}.\end{aligned}}}As a corollary, the product or quotient of tangents can be converted to a quotient of sums of cosines or sines, respectively,tan⁡θtan⁡φ=cos⁡(θ−φ)−cos⁡(θ+φ)cos⁡(θ−φ)+cos⁡(θ+φ),tan⁡θtan⁡φ=sin⁡(θ+φ)+sin⁡(θ−φ)sin⁡(θ+φ)−sin⁡(θ−φ).{\displaystyle {\begin{aligned}\tan \theta \,\tan \varphi &={\frac {\cos(\theta -\varphi )-\cos(\theta +\varphi )}{\cos(\theta -\varphi )+\cos(\theta +\varphi )}},\\[5mu]{\frac {\tan \theta }{\tan \varphi }}&={\frac {\sin(\theta +\varphi )+\sin(\theta -\varphi )}{\sin(\theta +\varphi )-\sin(\theta -\varphi )}}.\end{aligned}}}

More generally, for a product of any number of sines or cosines,[citation needed]∏k=1ncos⁡θk=12n∑e∈Scos⁡(e1θ1+⋯+enθn)where e=(e1,…,en)∈S={1,−1}n,∏k=1nsin⁡θk=(−1)⌊n2⌋2n{∑e∈Scos⁡(e1θ1+⋯+enθn)∏j=1nejifnis even,∑e∈Ssin⁡(e1θ1+⋯+enθn)∏j=1nejifnis odd.{\displaystyle {\begin{aligned}\prod _{k=1}^{n}\cos \theta _{k}&={\frac {1}{2^{n}}}\sum _{e\in S}\cos(e_{1}\theta _{1}+\cdots +e_{n}\theta _{n})\\[5mu]&{\text{where }}e=(e_{1},\ldots ,e_{n})\in S=\{1,-1\}^{n},\\\prod _{k=1}^{n}\sin \theta _{k}&={\frac {(-1)^{\left\lfloor {\frac {n}{2}}\right\rfloor }}{2^{n}}}{\begin{cases}\displaystyle \sum _{e\in S}\cos(e_{1}\theta _{1}+\cdots +e_{n}\theta _{n})\prod _{j=1}^{n}e_{j}\;{\text{if}}\;n\;{\text{is even}},\\\displaystyle \sum _{e\in S}\sin(e_{1}\theta _{1}+\cdots +e_{n}\theta _{n})\prod _{j=1}^{n}e_{j}\;{\text{if}}\;n\;{\text{is odd}}.\end{cases}}\end{aligned}}}

Sum-to-product identities

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Diagram illustrating sum-to-product identities for sine and cosine. The blue right-angled triangle has angleθ{\displaystyle \theta } and the red right-angled triangle has angleφ{\displaystyle \varphi }. Both have a hypotenuse of length 1. Auxiliary angles, here calledp{\displaystyle p} andq{\displaystyle q}, are constructed such thatp=12(θ+φ){\displaystyle p={\tfrac {1}{2}}(\theta +\varphi )} andq=12(θ−φ){\displaystyle q={\tfrac {1}{2}}(\theta -\varphi )}. Therefore,θ=p+q{\displaystyle \theta =p+q} andφ=p−q{\displaystyle \varphi =p-q}. This allows the two congruent purple-outline trianglesAFG{\displaystyle AFG} andFCE{\displaystyle FCE} to be constructed, each with hypotenusecos⁡q{\displaystyle \cos q} and anglep{\displaystyle p} at their base. The sum of the heights of the red and blue triangles issin⁡θ+sin⁡φ{\displaystyle \sin \theta +\sin \varphi }, and this is equal to twice the height of one purple triangle, i.e.2sin⁡pcos⁡q{\displaystyle 2\sin p\cos q}. Writingp{\displaystyle p} andq{\displaystyle q} in that equation in terms ofθ{\displaystyle \theta } andφ{\displaystyle \varphi } yields a sum-to-product identity for sine:sin⁡θ+sin⁡φ=2sin⁡12(θ+φ)cos⁡12(θ−φ){\displaystyle \sin \theta +\sin \varphi =2\sin {\tfrac {1}{2}}(\theta +\varphi )\,\cos {\tfrac {1}{2}}(\theta -\varphi )}. Similarly, the sum of the widths of the red and blue triangles yields the corresponding identity for cosine.

The sum of sines or cosines of two angles can be converted to a product of sines or cosines of the mean and half the difference of the angles:[32]

sin⁡θ+sin⁡φ=2sin⁡12(θ+φ)cos⁡12(θ−φ),sin⁡θ−sin⁡φ=2cos⁡12(θ+φ)sin⁡12(θ−φ),cos⁡θ+cos⁡φ=2cos⁡12(θ+φ)cos⁡12(θ−φ),cos⁡θ−cos⁡φ=−2sin⁡12(θ+φ)sin⁡12(θ−φ).{\displaystyle {\begin{aligned}\sin \theta +\sin \varphi &=2\sin {\tfrac {1}{2}}(\theta +\varphi )\,\cos {\tfrac {1}{2}}(\theta -\varphi ),\\[5mu]\sin \theta -\sin \varphi &=2\cos {\tfrac {1}{2}}(\theta +\varphi )\,\sin {\tfrac {1}{2}}(\theta -\varphi ),\\[5mu]\cos \theta +\cos \varphi &=2\cos {\tfrac {1}{2}}(\theta +\varphi )\,\cos {\tfrac {1}{2}}(\theta -\varphi ),\\[5mu]\cos \theta -\cos \varphi &=-2\sin {\tfrac {1}{2}}(\theta +\varphi )\,\sin {\tfrac {1}{2}}(\theta -\varphi ).\end{aligned}}}

The sum of the tangent of two angles can be converted to a quotient of the sine of angles divided by the product of the cosines:[32]tan⁡θ±tan⁡φ=sin⁡(θ±φ)cos⁡θcos⁡φ.{\displaystyle \tan \theta \pm \tan \varphi ={\frac {\sin(\theta \pm \varphi )}{\cos \theta \,\cos \varphi }}.}

Hermite's cotangent identity

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Main article:Hermite's cotangent identity

Charles Hermite demonstrated the following identity.[33] Supposea1,…,an{\displaystyle a_{1},\ldots ,a_{n}} arecomplex numbers, no two of which differ by an integer multiple of π. Let

An,k=∏1≤j≤nj≠kcot⁡(ak−aj){\displaystyle A_{n,k}=\prod _{\begin{smallmatrix}1\leq j\leq n\\j\neq k\end{smallmatrix}}\cot(a_{k}-a_{j})}

(in particular,A1,1,{\displaystyle A_{1,1},} being anempty product, is 1). Then

cot⁡(z−a1)⋯cot⁡(z−an)=cos⁡nπ2+∑k=1nAn,kcot⁡(z−ak).{\displaystyle \cot(z-a_{1})\cdots \cot(z-a_{n})=\cos {\frac {n\pi }{2}}+\sum _{k=1}^{n}A_{n,k}\cot(z-a_{k}).}

The simplest non-trivial example is the case n = 2:

cot⁡(z−a1)cot⁡(z−a2)=−1+cot⁡(a1−a2)cot⁡(z−a1)+cot⁡(a2−a1)cot⁡(z−a2).{\displaystyle \cot(z-a_{1})\cot(z-a_{2})=-1+\cot(a_{1}-a_{2})\cot(z-a_{1})+\cot(a_{2}-a_{1})\cot(z-a_{2}).}

Finite products of trigonometric functions

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Forcoprime integersn,m

∏k=1n(2a+2cos⁡(2πkmn+x))=2(Tn(a)+(−1)n+mcos⁡(nx)){\displaystyle \prod _{k=1}^{n}\left(2a+2\cos \left({\frac {2\pi km}{n}}+x\right)\right)=2\left(T_{n}(a)+{(-1)}^{n+m}\cos(nx)\right)}

whereTn is theChebyshev polynomial.[citation needed]

The following relationship holds for the sine function

∏k=1n−1sin⁡(kπn)=n2n−1.{\displaystyle \prod _{k=1}^{n-1}\sin \left({\frac {k\pi }{n}}\right)={\frac {n}{2^{n-1}}}.}

More generally for an integern > 0[34]

sin⁡(nx)=2n−1∏k=0n−1sin⁡(knπ+x)=2n−1∏k=1nsin⁡(knπ−x).{\displaystyle \sin(nx)=2^{n-1}\prod _{k=0}^{n-1}\sin \left({\frac {k}{n}}\pi +x\right)=2^{n-1}\prod _{k=1}^{n}\sin \left({\frac {k}{n}}\pi -x\right).}

or written in terms of thechord functioncrd⁡x≡2sin⁡12x{\textstyle \operatorname {crd} x\equiv 2\sin {\tfrac {1}{2}}x},

crd⁡(nx)=∏k=1ncrd⁡(kn2π−x).{\displaystyle \operatorname {crd} (nx)=\prod _{k=1}^{n}\operatorname {crd} \left({\frac {k}{n}}2\pi -x\right).}

This comes from thefactorization of the polynomialzn−1{\textstyle z^{n}-1} into linear factors (cf.root of unity): For any complexz and an integern > 0,

zn−1=∏k=1n(z−exp⁡(kn2πi)).{\displaystyle z^{n}-1=\prod _{k=1}^{n}\left(z-\exp {\Bigl (}{\frac {k}{n}}2\pi i{\Bigr )}\right).}

Linear combinations

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For some purposes it is important to know that anylinear combination of sine waves of the same period or frequency but differentphase shifts is also a sine wave with the same period or frequency, but a different phase shift. This is useful insinusoiddata fitting, because the measured or observed data are linearly related to thea andb unknowns of thein-phase and quadrature components basis below, resulting in a simplerJacobian, compared to that ofc{\displaystyle c} andφ{\displaystyle \varphi }.

Sine and cosine

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The linear combination, or harmonic addition, of sine and cosine waves is equivalent to a single sine wave with a phase shift and scaled amplitude,[35][36]

acos⁡x+bsin⁡x=ccos⁡(x+φ){\displaystyle a\cos x+b\sin x=c\cos(x+\varphi )}

wherec{\displaystyle c} andφ{\displaystyle \varphi } are defined as so:

c=sgn⁡(a)a2+b2,φ=arctan(−b/a),{\displaystyle {\begin{aligned}c&=\operatorname {sgn}(a){\sqrt {a^{2}+b^{2}}},\\\varphi &={\arctan }{\bigl (}{-b/a}{\bigr )},\end{aligned}}}

given thata≠0.{\displaystyle a\neq 0.}

Arbitrary phase shift

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More generally, for arbitrary phase shifts, we have

asin⁡(x+θa)+bsin⁡(x+θb)=csin⁡(x+φ){\displaystyle a\sin(x+\theta _{a})+b\sin(x+\theta _{b})=c\sin(x+\varphi )}

wherec{\displaystyle c} andφ{\displaystyle \varphi } satisfy:

c2=a2+b2+2abcos⁡(θa−θb),tan⁡φ=asin⁡θa+bsin⁡θbacos⁡θa+bcos⁡θb.{\displaystyle {\begin{aligned}c^{2}&=a^{2}+b^{2}+2ab\cos \left(\theta _{a}-\theta _{b}\right),\\\tan \varphi &={\frac {a\sin \theta _{a}+b\sin \theta _{b}}{a\cos \theta _{a}+b\cos \theta _{b}}}.\end{aligned}}}

More than two sinusoids

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See also:Phasor addition

The general case reads[36]

∑iaisin⁡(x+θi)=asin⁡(x+θ),{\displaystyle \sum _{i}a_{i}\sin(x+\theta _{i})=a\sin(x+\theta ),}wherea2=∑i,jaiajcos⁡(θi−θj){\displaystyle a^{2}=\sum _{i,j}a_{i}a_{j}\cos(\theta _{i}-\theta _{j})}andtan⁡θ=∑iaisin⁡θi∑iaicos⁡θi.{\displaystyle \tan \theta ={\frac {\sum _{i}a_{i}\sin \theta _{i}}{\sum _{i}a_{i}\cos \theta _{i}}}.}

Lagrange's trigonometric identities

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These identities, named afterJoseph Louis Lagrange, are:[37][38][39]∑k=0nsin⁡kθ=cos⁡12θ−cos⁡((n+12)θ)2sin⁡12θ∑k=1ncos⁡kθ=−sin⁡12θ+sin⁡((n+12)θ)2sin⁡12θ{\displaystyle {\begin{aligned}\sum _{k=0}^{n}\sin k\theta &={\frac {\cos {\tfrac {1}{2}}\theta -\cos \left(\left(n+{\tfrac {1}{2}}\right)\theta \right)}{2\sin {\tfrac {1}{2}}\theta }}\\[5pt]\sum _{k=1}^{n}\cos k\theta &={\frac {-\sin {\tfrac {1}{2}}\theta +\sin \left(\left(n+{\tfrac {1}{2}}\right)\theta \right)}{2\sin {\tfrac {1}{2}}\theta }}\end{aligned}}}forθ≢0(mod2π).{\displaystyle \theta \not \equiv 0{\pmod {2\pi }}.}

A related function is theDirichlet kernel:

Dn(θ)=1+2∑k=1ncos⁡kθ=sin⁡((n+12)θ)sin⁡12θ.{\displaystyle D_{n}(\theta )=1+2\sum _{k=1}^{n}\cos k\theta ={\frac {\sin \left(\left(n+{\tfrac {1}{2}}\right)\theta \right)}{\sin {\tfrac {1}{2}}\theta }}.}

A similar identity is[40]

∑k=1ncos⁡(2k−1)α=sin⁡(2nα)2sin⁡α.{\displaystyle \sum _{k=1}^{n}\cos(2k-1)\alpha ={\frac {\sin(2n\alpha )}{2\sin \alpha }}.}

The proof is the following. By using theangle sum and difference identities,sin⁡(A+B)−sin⁡(A−B)=2cos⁡Asin⁡B.{\displaystyle \sin(A+B)-\sin(A-B)=2\cos A\sin B.}Then let's examine the following formula,

2sin⁡α∑k=1ncos⁡(2k−1)α=2sin⁡αcos⁡α+2sin⁡αcos⁡3α+2sin⁡αcos⁡5α+⋯+2sin⁡αcos⁡(2n−1)α{\displaystyle 2\sin \alpha \sum _{k=1}^{n}\cos(2k-1)\alpha =2\sin \alpha \cos \alpha +2\sin \alpha \cos 3\alpha +2\sin \alpha \cos 5\alpha +\cdots +2\sin \alpha \cos(2n-1)\alpha }and this formula can be written by using the above identity,

2sin⁡α∑k=1ncos⁡(2k−1)α=∑k=1n(sin⁡(2kα)−sin⁡(2(k−1)α))=(sin⁡2α−sin⁡0)+(sin⁡4α−sin⁡2α)+(sin⁡6α−sin⁡4α)+⋯+(sin⁡(2nα)−sin⁡(2(n−1)α))=sin⁡(2nα).{\displaystyle {\begin{aligned}&2\sin \alpha \sum _{k=1}^{n}\cos(2k-1)\alpha \\&\quad =\sum _{k=1}^{n}(\sin(2k\alpha )-\sin(2(k-1)\alpha ))\\&\quad =(\sin 2\alpha -\sin 0)+(\sin 4\alpha -\sin 2\alpha )+(\sin 6\alpha -\sin 4\alpha )+\cdots +(\sin(2n\alpha )-\sin(2(n-1)\alpha ))\\&\quad =\sin(2n\alpha ).\end{aligned}}}

So, dividing this formula with2sin⁡α{\displaystyle 2\sin \alpha } completes the proof.

Certain linear fractional transformations

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Iff(x){\displaystyle f(x)} is given by thelinear fractional transformationf(x)=(cos⁡α)x−sin⁡α(sin⁡α)x+cos⁡α,{\displaystyle f(x)={\frac {(\cos \alpha )x-\sin \alpha }{(\sin \alpha )x+\cos \alpha }},}and similarlyg(x)=(cos⁡β)x−sin⁡β(sin⁡β)x+cos⁡β,{\displaystyle g(x)={\frac {(\cos \beta )x-\sin \beta }{(\sin \beta )x+\cos \beta }},}thenf(g(x))=g(f(x))=(cos⁡(α+β))x−sin⁡(α+β)(sin⁡(α+β))x+cos⁡(α+β).{\displaystyle f{\big (}g(x){\big )}=g{\big (}f(x){\big )}={\frac {{\big (}\cos(\alpha +\beta ){\big )}x-\sin(\alpha +\beta )}{{\big (}\sin(\alpha +\beta ){\big )}x+\cos(\alpha +\beta )}}.}

More tersely stated, if for allα{\displaystyle \alpha } we letfα{\displaystyle f_{\alpha }} be what we calledf{\displaystyle f} above, thenfα∘fβ=fα+β.{\displaystyle f_{\alpha }\circ f_{\beta }=f_{\alpha +\beta }.}

Ifx{\displaystyle x} is the slope of a line, thenf(x){\displaystyle f(x)} is the slope of its rotation through an angle of−α.{\displaystyle -\alpha .}

Relation to the complex exponential function

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Main article:Euler's formula

Euler's formula states that, for any real numberx:[41]eix=cos⁡x+isin⁡x,{\displaystyle e^{ix}=\cos x+i\sin x,}wherei is theimaginary unit. Substituting −x forx gives us:e−ix=cos⁡(−x)+isin⁡(−x)=cos⁡x−isin⁡x.{\displaystyle e^{-ix}=\cos(-x)+i\sin(-x)=\cos x-i\sin x.}

These two equations can be used to solve for cosine and sine in terms of theexponential function. Specifically,[42][43]cos⁡x=eix+e−ix2{\displaystyle \cos x={\frac {e^{ix}+e^{-ix}}{2}}}sin⁡x=eix−e−ix2i{\displaystyle \sin x={\frac {e^{ix}-e^{-ix}}{2i}}}

These formulae are useful for proving many other trigonometric identities. For example, thatei(θ+φ) =eiθeiφ means that

cos(θ +φ) +i sin(θ +φ) = (cosθ +i sinθ) (cosφ +i sinφ) = (cosθ cosφ − sinθ sinφ) +i (cosθ sinφ + sinθ cosφ).

That the real part of the left hand side equals the real part of the right hand side is an angle addition formula for cosine. The equality of the imaginary parts gives an angle addition formula for sine.

The following table expresses the trigonometric functions and their inverses in terms of the exponential function and thecomplex logarithm.

FunctionInverse function[44]
sin⁡θ=eiθ−e−iθ2i{\displaystyle \sin \theta ={\frac {e^{i\theta }-e^{-i\theta }}{2i}}}arcsin⁡x=−iln⁡(ix+1−x2){\displaystyle \arcsin x=-i\,\ln \left(ix+{\sqrt {1-x^{2}}}\right)}
cos⁡θ=eiθ+e−iθ2{\displaystyle \cos \theta ={\frac {e^{i\theta }+e^{-i\theta }}{2}}}arccos⁡x=−iln⁡(x+x2−1){\displaystyle \arccos x=-i\ln \left(x+{\sqrt {x^{2}-1}}\right)}
tan⁡θ=−ieiθ−e−iθeiθ+e−iθ{\displaystyle \tan \theta =-i\,{\frac {e^{i\theta }-e^{-i\theta }}{e^{i\theta }+e^{-i\theta }}}}arctan⁡x=i2ln⁡(i+xi−x){\displaystyle \arctan x={\frac {i}{2}}\ln \left({\frac {i+x}{i-x}}\right)}
csc⁡θ=2ieiθ−e−iθ{\displaystyle \csc \theta ={\frac {2i}{e^{i\theta }-e^{-i\theta }}}}arccsc⁡x=−iln⁡(ix+1−1x2){\displaystyle \operatorname {arccsc} x=-i\,\ln \left({\frac {i}{x}}+{\sqrt {1-{\frac {1}{x^{2}}}}}\right)}
sec⁡θ=2eiθ+e−iθ{\displaystyle \sec \theta ={\frac {2}{e^{i\theta }+e^{-i\theta }}}}arcsec⁡x=−iln⁡(1x+i1−1x2){\displaystyle \operatorname {arcsec} x=-i\,\ln \left({\frac {1}{x}}+i{\sqrt {1-{\frac {1}{x^{2}}}}}\right)}
cot⁡θ=ieiθ+e−iθeiθ−e−iθ{\displaystyle \cot \theta =i\,{\frac {e^{i\theta }+e^{-i\theta }}{e^{i\theta }-e^{-i\theta }}}}arccot⁡x=i2ln⁡(x−ix+i){\displaystyle \operatorname {arccot} x={\frac {i}{2}}\ln \left({\frac {x-i}{x+i}}\right)}
cis⁡θ=eiθ{\displaystyle \operatorname {cis} \theta =e^{i\theta }}arccis⁡x=−iln⁡x{\displaystyle \operatorname {arccis} x=-i\ln x}

Relation to complex hyperbolic functions

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Trigonometric functions may be deduced fromhyperbolic functions withcomplex arguments. The formulae for the relations are shown below[45][46].sin⁡x=−isinh⁡(ix)cos⁡x=cosh⁡(ix)tan⁡x=−itanh⁡(ix)cot⁡x=icoth⁡(ix)sec⁡x=sech⁡(ix)csc⁡x=icsch⁡(ix){\displaystyle {\begin{aligned}\sin x&=-i\sinh(ix)\\\cos x&=\cosh(ix)\\\tan x&=-i\tanh(ix)\\\cot x&=i\coth(ix)\\\sec x&=\operatorname {sech} (ix)\\\csc x&=i\operatorname {csch} (ix)\\\end{aligned}}}

Series expansion

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When using apower series expansion to define trigonometric functions, the following identities are obtained:[47]

sin⁡x=x−x33!+x55!−x77!+⋯=∑n=0∞(−1)nx2n+1(2n+1)!cos⁡x=1−x22!+x44!−x66!+⋯=∑n=0∞(−1)nx2n(2n)!{\displaystyle {\begin{aligned}\sin x&=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots &&=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n+1}}{(2n+1)!}}\\\cos x&=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+\cdots &&=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n}}{(2n)!}}\end{aligned}}}

Infinite product formulae

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For applications tospecial functions, the followinginfinite product formulae for trigonometric functions are useful:[48][49]

sin⁡x=x∏n=1∞(1−x2π2n2),cos⁡x=∏n=1∞(1−x2π2(n−12))2),sinh⁡x=x∏n=1∞(1+x2π2n2),cosh⁡x=∏n=1∞(1+x2π2(n−12))2).{\displaystyle {\begin{aligned}\sin x&=x\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{\pi ^{2}n^{2}}}\right),&\cos x&=\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{\pi ^{2}\left(n-{\frac {1}{2}}\right)\!{\vphantom {)}}^{2}}}\right),\\[10mu]\sinh x&=x\prod _{n=1}^{\infty }\left(1+{\frac {x^{2}}{\pi ^{2}n^{2}}}\right),&\cosh x&=\prod _{n=1}^{\infty }\left(1+{\frac {x^{2}}{\pi ^{2}\left(n-{\frac {1}{2}}\right)\!{\vphantom {)}}^{2}}}\right).\end{aligned}}}

Inverse trigonometric functions

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Main article:Inverse trigonometric functions

The following identities give the result of composing a trigonometric function with an inverse trigonometric function.[50]

sin⁡(arcsin⁡x)=xcos⁡(arcsin⁡x)=1−x2tan⁡(arcsin⁡x)=x1−x2sin⁡(arccos⁡x)=1−x2cos⁡(arccos⁡x)=xtan⁡(arccos⁡x)=1−x2xsin⁡(arctan⁡x)=x1+x2cos⁡(arctan⁡x)=11+x2tan⁡(arctan⁡x)=xsin⁡(arccsc⁡x)=1xcos⁡(arccsc⁡x)=1−1x2tan⁡(arccsc⁡x)=1x1−1x2sin⁡(arcsec⁡x)=1−1x2cos⁡(arcsec⁡x)=1xtan⁡(arcsec⁡x)=x1−1x2sin⁡(arccot⁡x)=11+x2cos⁡(arccot⁡x)=x1+x2tan⁡(arccot⁡x)=1x{\displaystyle {\begin{aligned}\sin(\arcsin x)&=x&\cos(\arcsin x)&={\sqrt {1-x^{2}}}&\tan(\arcsin x)&={\frac {x}{\sqrt {1-x^{2}}}}\\\sin(\arccos x)&={\sqrt {1-x^{2}}}&\cos(\arccos x)&=x&\tan(\arccos x)&={\frac {\sqrt {1-x^{2}}}{x}}\\\sin(\arctan x)&={\frac {x}{\sqrt {1+x^{2}}}}&\cos(\arctan x)&={\frac {1}{\sqrt {1+x^{2}}}}&\tan(\arctan x)&=x\\\sin(\operatorname {arccsc} x)&={\frac {1}{x}}&\cos(\operatorname {arccsc} x)&={\sqrt {1-{\frac {1}{x^{2}}}}}&\tan(\operatorname {arccsc} x)&={\frac {1}{x{\sqrt {1-{\frac {1}{x^{2}}}}}}}\\\sin(\operatorname {arcsec} x)&={\sqrt {1-{\frac {1}{x^{2}}}}}&\cos(\operatorname {arcsec} x)&={\frac {1}{x}}&\tan(\operatorname {arcsec} x)&=x{\sqrt {1-{\frac {1}{x^{2}}}}}\\\sin(\operatorname {arccot} x)&={\frac {1}{\sqrt {1+x^{2}}}}&\cos(\operatorname {arccot} x)&={\frac {x}{\sqrt {1+x^{2}}}}&\tan(\operatorname {arccot} x)&={\frac {1}{x}}\\\end{aligned}}}

Taking themultiplicative inverse of both sides of the each equation above results in the equations forcsc=1sin,sec=1cos, and cot=1tan.{\displaystyle \csc ={\frac {1}{\sin }},\;\sec ={\frac {1}{\cos }},{\text{ and }}\cot ={\frac {1}{\tan }}.}The right hand side of the formula above will always be flipped.For example, the equation forcot⁡(arcsin⁡x){\displaystyle \cot(\arcsin x)} is:cot⁡(arcsin⁡x)=1tan⁡(arcsin⁡x)=1x1−x2=1−x2x{\displaystyle \cot(\arcsin x)={\frac {1}{\tan(\arcsin x)}}={\frac {1}{\frac {x}{\sqrt {1-x^{2}}}}}={\frac {\sqrt {1-x^{2}}}{x}}}while the equations forcsc⁡(arccos⁡x){\displaystyle \csc(\arccos x)} andsec⁡(arccos⁡x){\displaystyle \sec(\arccos x)} are:csc⁡(arccos⁡x)=1sin⁡(arccos⁡x)=11−x2 and sec⁡(arccos⁡x)=1cos⁡(arccos⁡x)=1x.{\displaystyle \csc(\arccos x)={\frac {1}{\sin(\arccos x)}}={\frac {1}{\sqrt {1-x^{2}}}}\qquad {\text{ and }}\quad \sec(\arccos x)={\frac {1}{\cos(\arccos x)}}={\frac {1}{x}}.}

The following identities are implied by thereflection identities. They hold wheneverx,r,s,−x,−r,{\displaystyle x,r,s,-x,-r,} and−s{\displaystyle -s} are in the domains of the relevant functions.π2 = arcsin⁡(x)+arccos⁡(x) = arctan⁡(r)+arccot⁡(r) = arcsec⁡(s)+arccsc⁡(s)π = arccos⁡(x)+arccos⁡(−x) = arccot⁡(r)+arccot⁡(−r) = arcsec⁡(s)+arcsec⁡(−s)0 = arcsin⁡(x)+arcsin⁡(−x) = arctan⁡(r)+arctan⁡(−r) = arccsc⁡(s)+arccsc⁡(−s){\displaystyle {\begin{alignedat}{9}{\frac {\pi }{2}}~&=~\arcsin(x)&&+\arccos(x)~&&=~\arctan(r)&&+\operatorname {arccot}(r)~&&=~\operatorname {arcsec}(s)&&+\operatorname {arccsc}(s)\\[0.4ex]\pi ~&=~\arccos(x)&&+\arccos(-x)~&&=~\operatorname {arccot}(r)&&+\operatorname {arccot}(-r)~&&=~\operatorname {arcsec}(s)&&+\operatorname {arcsec}(-s)\\[0.4ex]0~&=~\arcsin(x)&&+\arcsin(-x)~&&=~\arctan(r)&&+\arctan(-r)~&&=~\operatorname {arccsc}(s)&&+\operatorname {arccsc}(-s)\\[1.0ex]\end{alignedat}}}

Also,[51]arctan⁡x+arctan⁡1x={π2,if x>0−π2,if x<0arccot⁡x+arccot⁡1x={π2,if x>03π2,if x<0{\displaystyle {\begin{aligned}\arctan x+\arctan {\dfrac {1}{x}}&={\begin{cases}{\frac {\pi }{2}},&{\text{if }}x>0\\-{\frac {\pi }{2}},&{\text{if }}x<0\end{cases}}\\\operatorname {arccot} x+\operatorname {arccot} {\dfrac {1}{x}}&={\begin{cases}{\frac {\pi }{2}},&{\text{if }}x>0\\{\frac {3\pi }{2}},&{\text{if }}x<0\end{cases}}\\\end{aligned}}}arccos⁡1x=arcsec⁡x and arcsec⁡1x=arccos⁡x{\displaystyle \arccos {\frac {1}{x}}=\operatorname {arcsec} x\qquad {\text{ and }}\qquad \operatorname {arcsec} {\frac {1}{x}}=\arccos x}arcsin⁡1x=arccsc⁡x and arccsc⁡1x=arcsin⁡x{\displaystyle \arcsin {\frac {1}{x}}=\operatorname {arccsc} x\qquad {\text{ and }}\qquad \operatorname {arccsc} {\frac {1}{x}}=\arcsin x}

Thearctangent function can be expanded as a series:[52]arctan⁡(nx)=∑m=1narctan⁡x1+(m−1)mx2{\displaystyle \arctan(nx)=\sum _{m=1}^{n}\arctan {\frac {x}{1+(m-1)mx^{2}}}}

Identities without variables

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In terms of thearctangent function we have[51]arctan⁡12=arctan⁡13+arctan⁡17.{\displaystyle \arctan {\frac {1}{2}}=\arctan {\frac {1}{3}}+\arctan {\frac {1}{7}}.}

The curious identity known asMorrie's law,cos⁡20∘⋅cos⁡40∘⋅cos⁡80∘=18,{\displaystyle \cos 20^{\circ }\cdot \cos 40^{\circ }\cdot \cos 80^{\circ }={\frac {1}{8}},}

is a special case of an identity that contains one variable:∏j=0k−1cos⁡(2jx)=sin⁡(2kx)2ksin⁡x.{\displaystyle \prod _{j=0}^{k-1}\cos \left(2^{j}x\right)={\frac {\sin \left(2^{k}x\right)}{2^{k}\sin x}}.}

Similarly,sin⁡20∘⋅sin⁡40∘⋅sin⁡80∘=38{\displaystyle \sin 20^{\circ }\cdot \sin 40^{\circ }\cdot \sin 80^{\circ }={\frac {\sqrt {3}}{8}}}is a special case of an identity withx=20∘{\displaystyle x=20^{\circ }}:sin⁡x⋅sin⁡(60∘−x)⋅sin⁡(60∘+x)=sin⁡3x4.{\displaystyle \sin x\cdot \sin \left(60^{\circ }-x\right)\cdot \sin \left(60^{\circ }+x\right)={\frac {\sin 3x}{4}}.}

For the casex=15∘{\displaystyle x=15^{\circ }},sin⁡15∘⋅sin⁡45∘⋅sin⁡75∘=28,sin⁡15∘⋅sin⁡75∘=14.{\displaystyle {\begin{aligned}\sin 15^{\circ }\cdot \sin 45^{\circ }\cdot \sin 75^{\circ }&={\frac {\sqrt {2}}{8}},\\\sin 15^{\circ }\cdot \sin 75^{\circ }&={\frac {1}{4}}.\end{aligned}}}

For the casex=10∘{\displaystyle x=10^{\circ }},sin⁡10∘⋅sin⁡50∘⋅sin⁡70∘=18.{\displaystyle \sin 10^{\circ }\cdot \sin 50^{\circ }\cdot \sin 70^{\circ }={\frac {1}{8}}.}

The same cosine identity iscos⁡x⋅cos⁡(60∘−x)⋅cos⁡(60∘+x)=cos⁡3x4.{\displaystyle \cos x\cdot \cos \left(60^{\circ }-x\right)\cdot \cos \left(60^{\circ }+x\right)={\frac {\cos 3x}{4}}.}

Similarly,cos⁡10∘⋅cos⁡50∘⋅cos⁡70∘=38,cos⁡15∘⋅cos⁡45∘⋅cos⁡75∘=28,cos⁡15∘⋅cos⁡75∘=14.{\displaystyle {\begin{aligned}\cos 10^{\circ }\cdot \cos 50^{\circ }\cdot \cos 70^{\circ }&={\frac {\sqrt {3}}{8}},\\\cos 15^{\circ }\cdot \cos 45^{\circ }\cdot \cos 75^{\circ }&={\frac {\sqrt {2}}{8}},\\\cos 15^{\circ }\cdot \cos 75^{\circ }&={\frac {1}{4}}.\end{aligned}}}

Similarly,tan⁡50∘⋅tan⁡60∘⋅tan⁡70∘=tan⁡80∘,tan⁡40∘⋅tan⁡30∘⋅tan⁡20∘=tan⁡10∘.{\displaystyle {\begin{aligned}\tan 50^{\circ }\cdot \tan 60^{\circ }\cdot \tan 70^{\circ }&=\tan 80^{\circ },\\\tan 40^{\circ }\cdot \tan 30^{\circ }\cdot \tan 20^{\circ }&=\tan 10^{\circ }.\end{aligned}}}

The following is perhaps not as readily generalized to an identity containing variables (but see explanation below):cos⁡24∘+cos⁡48∘+cos⁡96∘+cos⁡168∘=12.{\displaystyle \cos 24^{\circ }+\cos 48^{\circ }+\cos 96^{\circ }+\cos 168^{\circ }={\frac {1}{2}}.}

Degree measure ceases to be more felicitous than radian measure when we consider this identity with 21 in the denominators:cos⁡2π21+cos⁡(2⋅2π21)+cos⁡(4⋅2π21)+cos⁡(5⋅2π21)+cos⁡(8⋅2π21)+cos⁡(10⋅2π21)=12.{\displaystyle \cos {\frac {2\pi }{21}}+\cos \left(2\cdot {\frac {2\pi }{21}}\right)+\cos \left(4\cdot {\frac {2\pi }{21}}\right)+\cos \left(5\cdot {\frac {2\pi }{21}}\right)+\cos \left(8\cdot {\frac {2\pi }{21}}\right)+\cos \left(10\cdot {\frac {2\pi }{21}}\right)={\frac {1}{2}}.}

The factors 1, 2, 4, 5, 8, 10 may start to make the pattern clear: they are those integers less than⁠21/2⁠ that arerelatively prime to (or have noprime factors in common with) 21. The last several examples are corollaries of a basic fact about the irreduciblecyclotomic polynomials: the cosines are the real parts of the zeroes of those polynomials; the sum of the zeroes is theMöbius function evaluated at (in the very last case above) 21; only half of the zeroes are present above. The two identities preceding this last one arise in the same fashion with 21 replaced by 10 and 15, respectively.

Other cosine identities include:[53]2cos⁡π3=1,2cos⁡π5×2cos⁡2π5=1,2cos⁡π7×2cos⁡2π7×2cos⁡3π7=1,{\displaystyle {\begin{aligned}2\cos {\frac {\pi }{3}}&=1,\\2\cos {\frac {\pi }{5}}\times 2\cos {\frac {2\pi }{5}}&=1,\\2\cos {\frac {\pi }{7}}\times 2\cos {\frac {2\pi }{7}}\times 2\cos {\frac {3\pi }{7}}&=1,\end{aligned}}}and so forth for all odd numbers, and hencecos⁡π3+cos⁡π5×cos⁡2π5+cos⁡π7×cos⁡2π7×cos⁡3π7+⋯=1.{\displaystyle \cos {\frac {\pi }{3}}+\cos {\frac {\pi }{5}}\times \cos {\frac {2\pi }{5}}+\cos {\frac {\pi }{7}}\times \cos {\frac {2\pi }{7}}\times \cos {\frac {3\pi }{7}}+\dots =1.}

Many of those curious identities stem from more general facts like the following:[54]∏k=1n−1sin⁡kπn=n2n−1{\displaystyle \prod _{k=1}^{n-1}\sin {\frac {k\pi }{n}}={\frac {n}{2^{n-1}}}}and∏k=1n−1cos⁡kπn=sin⁡πn22n−1.{\displaystyle \prod _{k=1}^{n-1}\cos {\frac {k\pi }{n}}={\frac {\sin {\frac {\pi n}{2}}}{2^{n-1}}}.}

Combining these gives us∏k=1n−1tan⁡kπn=nsin⁡πn2{\displaystyle \prod _{k=1}^{n-1}\tan {\frac {k\pi }{n}}={\frac {n}{\sin {\frac {\pi n}{2}}}}}

Ifn is an odd number (n=2m+1{\displaystyle n=2m+1}) we can make use of the symmetries to get∏k=1mtan⁡kπ2m+1=2m+1{\displaystyle \prod _{k=1}^{m}\tan {\frac {k\pi }{2m+1}}={\sqrt {2m+1}}}

The transfer function of theButterworth low pass filter can be expressed in terms of polynomial and poles. By setting the frequency as thecutoff frequency, the following identity can be proved:∏k=1nsin⁡(2k−1)π4n=∏k=1ncos⁡(2k−1)π4n=22n{\displaystyle \prod _{k=1}^{n}\sin {\frac {\left(2k-1\right)\pi }{4n}}=\prod _{k=1}^{n}\cos {\frac {\left(2k-1\right)\pi }{4n}}={\frac {\sqrt {2}}{2^{n}}}}

Computingπ

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An efficient way tocomputeπ to alarge number of digits is based on the following identity without variables, due toMachin. This is known as aMachin-like formula:π4=4arctan⁡15−arctan⁡1239{\displaystyle {\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}}or, alternatively, by using an identity ofLeonhard Euler:π4=5arctan⁡17+2arctan⁡379{\displaystyle {\frac {\pi }{4}}=5\arctan {\frac {1}{7}}+2\arctan {\frac {3}{79}}}or by usingPythagorean triples:π=arccos⁡45+arccos⁡513+arccos⁡1665=arcsin⁡35+arcsin⁡1213+arcsin⁡6365.{\displaystyle \pi =\arccos {\frac {4}{5}}+\arccos {\frac {5}{13}}+\arccos {\frac {16}{65}}=\arcsin {\frac {3}{5}}+\arcsin {\frac {12}{13}}+\arcsin {\frac {63}{65}}.}

Others include:[55][51]π4=arctan⁡12+arctan⁡13,{\displaystyle {\frac {\pi }{4}}=\arctan {\frac {1}{2}}+\arctan {\frac {1}{3}},}π=arctan⁡1+arctan⁡2+arctan⁡3,{\displaystyle \pi =\arctan 1+\arctan 2+\arctan 3,}π4=2arctan⁡13+arctan⁡17.{\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{3}}+\arctan {\frac {1}{7}}.}

Generally, for numberst1, ...,tn−1 ∈ (−1, 1) for whichθn = Σn−1
k=1 arctantk ∈ (π/4, 3π/4), lettn = tan(π/2 −θn) = cotθn. This last expression can be computed directly using the formula for the cotangent of a sum of angles whose tangents aret1, ...,tn−1 and its value will be in(−1, 1). In particular, the computedtn will be rational whenever all thet1, ...,tn−1 values are rational. With these values,π2=∑k=1narctan⁡(tk)π=∑k=1nsgn⁡(tk)arccos⁡(1−tk21+tk2)π=∑k=1narcsin⁡(2tk1+tk2)π=∑k=1narctan⁡(2tk1−tk2),{\displaystyle {\begin{aligned}{\frac {\pi }{2}}&=\sum _{k=1}^{n}\arctan(t_{k})\\\pi &=\sum _{k=1}^{n}\operatorname {sgn}(t_{k})\arccos \left({\frac {1-t_{k}^{2}}{1+t_{k}^{2}}}\right)\\\pi &=\sum _{k=1}^{n}\arcsin \left({\frac {2t_{k}}{1+t_{k}^{2}}}\right)\\\pi &=\sum _{k=1}^{n}\arctan \left({\frac {2t_{k}}{1-t_{k}^{2}}}\right)\,,\end{aligned}}}

where in all but the first expression, we have used tangent half-angle formulae. The first two formulae work even if one or more of thetk values is not within(−1, 1). Note that ift =p/q is rational, then the(2t, 1 −t2, 1 +t2) values in the above formulae are proportional to the Pythagorean triple(2pq,q2 −p2,q2 +p2).

For example, forn = 3 terms,π2=arctan⁡(ab)+arctan⁡(cd)+arctan⁡(bd−acad+bc){\displaystyle {\frac {\pi }{2}}=\arctan \left({\frac {a}{b}}\right)+\arctan \left({\frac {c}{d}}\right)+\arctan \left({\frac {bd-ac}{ad+bc}}\right)}for anya,b,c,d > 0.

An identity of Euclid

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Euclid showed in Book XIII, Proposition 10 of hisElements that the area of the square on the side of a regular pentagon inscribed in a circle is equal to the sum of the areas of the squares on the sides of the regular hexagon and the regular decagon inscribed in the same circle. In the language of modern trigonometry, this says:sin2⁡18∘+sin2⁡30∘=sin2⁡36∘.{\displaystyle \sin ^{2}18^{\circ }+\sin ^{2}30^{\circ }=\sin ^{2}36^{\circ }.}

Ptolemy used this proposition to compute some angles inhis table of chords in Book I, chapter 11 ofAlmagest.

Composition of trigonometric functions

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These identities involve a trigonometric function of a trigonometric function:[56]

cos⁡(tsin⁡x)=J0(t)+2∑k=1∞J2k(t)cos⁡(2kx){\displaystyle \cos(t\sin x)=J_{0}(t)+2\sum _{k=1}^{\infty }J_{2k}(t)\cos(2kx)}
sin⁡(tsin⁡x)=2∑k=0∞J2k+1(t)sin⁡((2k+1)x){\displaystyle \sin(t\sin x)=2\sum _{k=0}^{\infty }J_{2k+1}(t)\sin {\big (}(2k+1)x{\big )}}
cos⁡(tcos⁡x)=J0(t)+2∑k=1∞(−1)kJ2k(t)cos⁡(2kx){\displaystyle \cos(t\cos x)=J_{0}(t)+2\sum _{k=1}^{\infty }(-1)^{k}J_{2k}(t)\cos(2kx)}
sin⁡(tcos⁡x)=2∑k=0∞(−1)kJ2k+1(t)cos⁡((2k+1)x){\displaystyle \sin(t\cos x)=2\sum _{k=0}^{\infty }(-1)^{k}J_{2k+1}(t)\cos {\big (}(2k+1)x{\big )}}

whereJi areBessel functions.

Further "conditional" identities for the caseα +β +γ = 180°

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Aconditional trigonometric identity is a trigonometric identity that holds if specified conditions on the arguments to the trigonometric functions are satisfied.[57] The following formulae apply to arbitrary plane triangles and follow fromα+β+γ=180∘,{\displaystyle \alpha +\beta +\gamma =180^{\circ },} as long as the functions occurring in the formulae are well-defined (the latter applies only to the formulae in which tangents and cotangents occur).[58]tan⁡α+tan⁡β+tan⁡γ=tan⁡αtan⁡βtan⁡γ1=cot⁡βcot⁡γ+cot⁡γcot⁡α+cot⁡αcot⁡βcot⁡(α2)+cot⁡(β2)+cot⁡(γ2)=cot⁡(α2)cot⁡(β2)cot⁡(γ2)1=tan⁡(β2)tan⁡(γ2)+tan⁡(γ2)tan⁡(α2)+tan⁡(α2)tan⁡(β2)sin⁡α+sin⁡β+sin⁡γ=4cos⁡(α2)cos⁡(β2)cos⁡(γ2)−sin⁡α+sin⁡β+sin⁡γ=4cos⁡(α2)sin⁡(β2)sin⁡(γ2)cos⁡α+cos⁡β+cos⁡γ=4sin⁡(α2)sin⁡(β2)sin⁡(γ2)+1−cos⁡α+cos⁡β+cos⁡γ=4sin⁡(α2)cos⁡(β2)cos⁡(γ2)−1sin⁡(2α)+sin⁡(2β)+sin⁡(2γ)=4sin⁡αsin⁡βsin⁡γ−sin⁡(2α)+sin⁡(2β)+sin⁡(2γ)=4sin⁡αcos⁡βcos⁡γcos⁡(2α)+cos⁡(2β)+cos⁡(2γ)=−4cos⁡αcos⁡βcos⁡γ−1−cos⁡(2α)+cos⁡(2β)+cos⁡(2γ)=−4cos⁡αsin⁡βsin⁡γ+1sin2⁡α+sin2⁡β+sin2⁡γ=2cos⁡αcos⁡βcos⁡γ+2−sin2⁡α+sin2⁡β+sin2⁡γ=2cos⁡αsin⁡βsin⁡γcos2⁡α+cos2⁡β+cos2⁡γ=−2cos⁡αcos⁡βcos⁡γ+1−cos2⁡α+cos2⁡β+cos2⁡γ=−2cos⁡αsin⁡βsin⁡γ+1sin2⁡(2α)+sin2⁡(2β)+sin2⁡(2γ)=−2cos⁡(2α)cos⁡(2β)cos⁡(2γ)+2cos2⁡(2α)+cos2⁡(2β)+cos2⁡(2γ)=2cos⁡(2α)cos⁡(2β)cos⁡(2γ)+11=sin2⁡(α2)+sin2⁡(β2)+sin2⁡(γ2)+2sin⁡(α2)sin⁡(β2)sin⁡(γ2){\displaystyle {\begin{aligned}\tan \alpha +\tan \beta +\tan \gamma &=\tan \alpha \tan \beta \tan \gamma \\1&=\cot \beta \cot \gamma +\cot \gamma \cot \alpha +\cot \alpha \cot \beta \\\cot \left({\frac {\alpha }{2}}\right)+\cot \left({\frac {\beta }{2}}\right)+\cot \left({\frac {\gamma }{2}}\right)&=\cot \left({\frac {\alpha }{2}}\right)\cot \left({\frac {\beta }{2}}\right)\cot \left({\frac {\gamma }{2}}\right)\\1&=\tan \left({\frac {\beta }{2}}\right)\tan \left({\frac {\gamma }{2}}\right)+\tan \left({\frac {\gamma }{2}}\right)\tan \left({\frac {\alpha }{2}}\right)+\tan \left({\frac {\alpha }{2}}\right)\tan \left({\frac {\beta }{2}}\right)\\\sin \alpha +\sin \beta +\sin \gamma &=4\cos \left({\frac {\alpha }{2}}\right)\cos \left({\frac {\beta }{2}}\right)\cos \left({\frac {\gamma }{2}}\right)\\-\sin \alpha +\sin \beta +\sin \gamma &=4\cos \left({\frac {\alpha }{2}}\right)\sin \left({\frac {\beta }{2}}\right)\sin \left({\frac {\gamma }{2}}\right)\\\cos \alpha +\cos \beta +\cos \gamma &=4\sin \left({\frac {\alpha }{2}}\right)\sin \left({\frac {\beta }{2}}\right)\sin \left({\frac {\gamma }{2}}\right)+1\\-\cos \alpha +\cos \beta +\cos \gamma &=4\sin \left({\frac {\alpha }{2}}\right)\cos \left({\frac {\beta }{2}}\right)\cos \left({\frac {\gamma }{2}}\right)-1\\\sin(2\alpha )+\sin(2\beta )+\sin(2\gamma )&=4\sin \alpha \sin \beta \sin \gamma \\-\sin(2\alpha )+\sin(2\beta )+\sin(2\gamma )&=4\sin \alpha \cos \beta \cos \gamma \\\cos(2\alpha )+\cos(2\beta )+\cos(2\gamma )&=-4\cos \alpha \cos \beta \cos \gamma -1\\-\cos(2\alpha )+\cos(2\beta )+\cos(2\gamma )&=-4\cos \alpha \sin \beta \sin \gamma +1\\\sin ^{2}\alpha +\sin ^{2}\beta +\sin ^{2}\gamma &=2\cos \alpha \cos \beta \cos \gamma +2\\-\sin ^{2}\alpha +\sin ^{2}\beta +\sin ^{2}\gamma &=2\cos \alpha \sin \beta \sin \gamma \\\cos ^{2}\alpha +\cos ^{2}\beta +\cos ^{2}\gamma &=-2\cos \alpha \cos \beta \cos \gamma +1\\-\cos ^{2}\alpha +\cos ^{2}\beta +\cos ^{2}\gamma &=-2\cos \alpha \sin \beta \sin \gamma +1\\\sin ^{2}(2\alpha )+\sin ^{2}(2\beta )+\sin ^{2}(2\gamma )&=-2\cos(2\alpha )\cos(2\beta )\cos(2\gamma )+2\\\cos ^{2}(2\alpha )+\cos ^{2}(2\beta )+\cos ^{2}(2\gamma )&=2\cos(2\alpha )\,\cos(2\beta )\,\cos(2\gamma )+1\\1&=\sin ^{2}\left({\frac {\alpha }{2}}\right)+\sin ^{2}\left({\frac {\beta }{2}}\right)+\sin ^{2}\left({\frac {\gamma }{2}}\right)+2\sin \left({\frac {\alpha }{2}}\right)\,\sin \left({\frac {\beta }{2}}\right)\,\sin \left({\frac {\gamma }{2}}\right)\end{aligned}}}

Historical shorthands

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Main articles:Versine andExsecant

Theversine,coversine,haversine, andexsecant were used in navigation. For example, thehaversine formula was used to calculate the distance between two points on a sphere. They are rarely used today.

Miscellaneous

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Dirichlet kernel

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Main article:Dirichlet kernel

TheDirichlet kernelDn(x) is the function occurring on both sides of the next identity:1+2cos⁡x+2cos⁡(2x)+2cos⁡(3x)+⋯+2cos⁡(nx)=sin⁡((n+12)x)sin⁡(12x).{\displaystyle 1+2\cos x+2\cos(2x)+2\cos(3x)+\cdots +2\cos(nx)={\frac {\sin \left(\left(n+{\frac {1}{2}}\right)x\right)}{\sin \left({\frac {1}{2}}x\right)}}.}

Theconvolution of anyintegrable function of period2π{\displaystyle 2\pi } with the Dirichlet kernel coincides with the function'sn{\displaystyle n}th-degree Fourier approximation. The same holds for anymeasure orgeneralized function.

Tangent half-angle substitution

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Main article:Tangent half-angle substitution

If we sett=tan⁡x2,{\displaystyle t=\tan {\frac {x}{2}},} then[59]sin⁡x=2t1+t2;cos⁡x=1−t21+t2;eix=1+it1−it;dx=2dt1+t2,{\displaystyle \sin x={\frac {2t}{1+t^{2}}};\qquad \cos x={\frac {1-t^{2}}{1+t^{2}}};\qquad e^{ix}={\frac {1+it}{1-it}};\qquad dx={\frac {2\,dt}{1+t^{2}}},}whereeix=cos⁡x+isin⁡x,{\displaystyle e^{ix}=\cos x+i\sin x,} sometimes abbreviated to cisx.

When this substitution oft{\displaystyle t} fortan⁠x/2⁠ is used incalculus, it follows thatsin⁡x{\displaystyle \sin x} is replaced by⁠2t/1 +t2⁠,cos⁡x{\displaystyle \cos x} is replaced by⁠1 −t2/1 +t2⁠ and the differentialdx is replaced by⁠2 dt/1 +t2⁠. Thereby one converts rational functions ofsin⁡x{\displaystyle \sin x} andcos⁡x{\displaystyle \cos x} to rational functions oft{\displaystyle t} in order to find theirantiderivatives.

Viète's infinite product

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See also:Viète's formula andSinc function

cos⁡θ2⋅cos⁡θ4⋅cos⁡θ8⋯=∏n=1∞cos⁡θ2n=sin⁡θθ=sinc⁡θ.{\displaystyle \cos {\frac {\theta }{2}}\cdot \cos {\frac {\theta }{4}}\cdot \cos {\frac {\theta }{8}}\cdots =\prod _{n=1}^{\infty }\cos {\frac {\theta }{2^{n}}}={\frac {\sin \theta }{\theta }}=\operatorname {sinc} \theta .}

See also

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References

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  1. ^Abramowitz, Milton;Stegun, Irene Ann, eds. (1983) [June 1964]."Chapter 4, eqn 4.3.45".Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. Applied Mathematics Series. Vol. 55 (Ninth reprint with additional corrections of tenth original printing with corrections (December 1972); first ed.). Washington D.C.; New York: United States Department of Commerce, National Bureau of Standards; Dover Publications. p. 73.ISBN 978-0-486-61272-0.LCCN 64-60036.MR 0167642.LCCN 65-12253.
  2. ^Selby 1970, p. 188
  3. ^Abramowitz and Stegun, p. 72, 4.3.13–15
  4. ^Abramowitz and Stegun, p. 72, 4.3.7–9
  5. ^Abramowitz and Stegun, p. 72, 4.3.16
  6. ^abcdWeisstein, Eric W."Trigonometric Addition Formulas".MathWorld.
  7. ^Abramowitz and Stegun, p. 72, 4.3.17
  8. ^Abramowitz and Stegun, p. 72, 4.3.18
  9. ^ab"Angle Sum and Difference Identities".www.milefoot.com. Retrieved2019-10-12.
  10. ^Abramowitz and Stegun, p. 72, 4.3.19
  11. ^Abramowitz and Stegun, p. 80, 4.4.32
  12. ^Abramowitz and Stegun, p. 80, 4.4.33
  13. ^Abramowitz and Stegun, p. 80, 4.4.34
  14. ^Bronstein, Manuel (1989). "Simplification of real elementary functions". In Gonnet, G. H. (ed.).Proceedings of the ACM-SIGSAM 1989 International Symposium on Symbolic and Algebraic Computation. ISSAC '89 (Portland US-OR, 1989-07). New York:ACM. pp. 207–211.doi:10.1145/74540.74566.ISBN 0-89791-325-6.
  15. ^Michael Hardy. (2016). "On Tangents and Secants of Infinite Sums."The American Mathematical Monthly, volume 123, number 7, 701–703.https://doi.org/10.4169/amer.math.monthly.123.7.701
  16. ^Michael Hardy (2025), "Invariance of the Cauchy Family Under Linear Fractional Transformations,"The American Mathematical Monthly, 132:5, 453–455, DOI: 10.1080/00029890.2025.2459048
  17. ^ Knight F. B., "A characterization of the Cauchy type."Proceedings of the American Mathematical Society, 1976:130–135.
  18. ^Hardy, Michael (2016)."On Tangents and Secants of Infinite Sums".American Mathematical Monthly.123 (7):701–703.doi:10.4169/amer.math.monthly.123.7.701.
  19. ^ab"Sine, Cosine, and Ptolemy's Theorem".
  20. ^abWeisstein, Eric W."Multiple-Angle Formulas".MathWorld.
  21. ^Abramowitz and Stegun, p. 74, 4.3.48
  22. ^abSelby 1970, pg. 190
  23. ^Weisstein, Eric W."Multiple-Angle Formulas".mathworld.wolfram.com. Retrieved2022-02-06.
  24. ^Ward, Ken."Multiple angles recursive formula".Ken Ward's Mathematics Pages.
  25. ^abAbramowitz, Milton;Stegun, Irene Ann, eds. (1983) [June 1964]."Chapter 4, eqn 4.3.20-22".Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. Applied Mathematics Series. Vol. 55 (Ninth reprint with additional corrections of tenth original printing with corrections (December 1972); first ed.). Washington D.C.; New York: United States Department of Commerce, National Bureau of Standards; Dover Publications. p. 72.ISBN 978-0-486-61272-0.LCCN 64-60036.MR 0167642.LCCN 65-12253.
  26. ^abWeisstein, Eric W."Half-Angle Formulas".MathWorld.
  27. ^Abramowitz and Stegun, p. 72, 4.3.24–26
  28. ^Weisstein, Eric W."Double-Angle Formulas".MathWorld.
  29. ^Abramowitz and Stegun, p. 72, 4.3.27–28
  30. ^Abramowitz and Stegun, p. 72, 4.3.31–33
  31. ^Eves, Howard (1990).An introduction to the history of mathematics (6th ed.). Philadelphia: Saunders College Pub. p. 309.ISBN 0-03-029558-0.OCLC 20842510.
  32. ^abAbramowitz and Stegun, p. 72, 4.3.34–39
  33. ^Johnson, Warren P. (Apr 2010). "Trigonometric Identities à la Hermite".American Mathematical Monthly.117 (4):311–327.doi:10.4169/000298910x480784.S2CID 29690311.
  34. ^"Product Identity Multiple Angle".
  35. ^Apostol, T.M. (1967) Calculus. 2nd edition. New York, NY, Wiley. Pp 334-335.
  36. ^abWeisstein, Eric W."Harmonic Addition Theorem".MathWorld.
  37. ^Ortiz Muñiz, Eddie (Feb 1953). "A Method for Deriving Various Formulas in Electrostatics and Electromagnetism Using Lagrange's Trigonometric Identities".American Journal of Physics.21 (2): 140.Bibcode:1953AmJPh..21..140M.doi:10.1119/1.1933371.
  38. ^Agarwal, Ravi P.; O'Regan, Donal (2008).Ordinary and Partial Differential Equations: With Special Functions, Fourier Series, and Boundary Value Problems (illustrated ed.). Springer Science & Business Media. p. 185.ISBN 978-0-387-79146-3.Extract of page 185
  39. ^Jeffrey, Alan; Dai, Hui-hui (2008). "Section 2.4.1.6".Handbook of Mathematical Formulas and Integrals (4th ed.). Academic Press.ISBN 978-0-12-374288-9.
  40. ^Fay, Temple H.; Kloppers, P. Hendrik (2001)."The Gibbs' phenomenon".International Journal of Mathematical Education in Science and Technology.32 (1):73–89.doi:10.1080/00207390117151.
  41. ^Abramowitz and Stegun, p. 74, 4.3.47
  42. ^Abramowitz and Stegun, p. 71, 4.3.2
  43. ^Abramowitz and Stegun, p. 71, 4.3.1
  44. ^Abramowitz and Stegun, p. 80, 4.4.26–31
  45. ^Hawkins, Faith Mary; Hawkins, J. Q. (March 1, 1969).Complex Numbers and Elementary Complex Functions. London: MacDonald Technical & Scientific London (published 1968). p. 122.ISBN 978-0356025056.
  46. ^Markushevich, A. I. (1966).The Remarkable Sine Function. New York: American Elsevier Publishing Company, Inc. pp. 35–37, 81.ISBN 978-1483256313.
  47. ^Abramowitz and Stegun, p. 74, 4.3.65–66
  48. ^Abramowitz and Stegun, p. 75, 4.3.89–90
  49. ^Abramowitz and Stegun, p. 85, 4.5.68–69
  50. ^Abramowitz & Stegun 1972, p. 73, 4.3.45
  51. ^abcWu, Rex H. "Proof Without Words: Euler's Arctangent Identity",Mathematics Magazine 77(3), June 2004, p. 189.
  52. ^S. M. Abrarov; R. K. Jagpal; R. Siddiqui; B. M. Quine (2021), "Algorithmic determination of a large integer in the two-term Machin-like formula for π",Mathematics,9 (17), 2162,arXiv:2107.01027,doi:10.3390/math9172162
  53. ^Humble, Steve (Nov 2004). "Grandma's identity".Mathematical Gazette.88:524–525.doi:10.1017/s0025557200176223.S2CID 125105552.
  54. ^Weisstein, Eric W."Sine".MathWorld.
  55. ^Harris, Edward M. "Sums of Arctangents", in Roger B. Nelson,Proofs Without Words (1993, Mathematical Association of America), p. 39.
  56. ^Milton Abramowitz and Irene Stegun,Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables,Dover Publications, New York, 1972, formulae 9.1.42–9.1.45
  57. ^Er. K. C. Joshi,Krishna's IIT MATHEMATIKA. Krishna Prakashan Media. Meerut, India. page 636.
  58. ^Cagnoli, Antonio (1808),Trigonométrie rectiligne et sphérique, p. 27.
  59. ^Abramowitz and Stegun, p. 72, 4.3.23

Bibliography

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External links

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