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Law of cosines

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From Wikipedia, the free encyclopedia

Generalization of Pythagorean theorem

This article is about the trigonometric identity. For the cosine law of optics, seeLambert's cosine law.

Fig. 1 – A triangle. The anglesα (orA),β (orB), andγ (orC) are respectively opposite the sidesa,b, andc.

Intrigonometry, thelaw of cosines (also known as thecosine formula orcosine rule) relates the lengths of the sides of atriangle to thecosine of one of itsangles. For a triangle with sides⁠ $a {\displaystyle a}$ ⁠,⁠ $b {\displaystyle b}$ ⁠, and⁠ $c {\displaystyle c}$ ⁠, opposite respective angles⁠ $\alpha$ ⁠,⁠ $\beta$ ⁠, and⁠ $\gamma$ ⁠ (see Fig. 1), the law of cosines states:

${\begin{aligned}c^{2}&=a^{2}+b^{2}-2ab\cos \gamma ,\\[3mu]a^{2}&=b^{2}+c^{2}-2bc\cos \alpha ,\\[3mu]b^{2}&=a^{2}+c^{2}-2ac\cos \beta .\end{aligned}}$

The law of cosines generalizes thePythagorean theorem, which holds only forright triangles: if⁠ $\gamma$ ⁠ is aright angle then⁠ $\cos \gamma =0$ ⁠, and the law of cosinesreduces to⁠ $c^{2}=a^{2}+b^{2}$ ⁠.

The law of cosines is useful forsolving a triangle when all three sides or two sides and their included angle are given.

Use in solving triangles

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Fig. 3 – Applications of the law of cosines: unknown side and unknown angle.

Given triangle sidesb andc and angleγ there are sometimes two solutions fora.

The theorem is used insolution of triangles, i.e., to find (see Figure 3):

the third side of a triangle if two sides and the angle between them is known: $c={\sqrt {a^{2}+b^{2}-2ab\cos \gamma }}\,;$
the angles of a triangle if the three sides are known: $\gamma =\arccos \left({\frac {a^{2}+b^{2}-c^{2}}{2ab}}\right)\,;$
the third side of a triangle if two sides and an angle opposite to one of them is known (this side can also be found by two applications of thelaw of sines):^[a] $a=b\cos \gamma \pm {\sqrt {c^{2}-b^{2}\sin ^{2}\gamma }}\,.$

These formulas produce highround-off errors infloating point calculations if the triangle is very acute, i.e., ifc is small relative toa andb orγ is small compared to 1. It is even possible to obtain a result slightly greater than one for the cosine of an angle.

The third formula shown is the result of solving fora in thequadratic equationa² − 2ab cosγ +b² −c² = 0. This equation can have 2, 1, or 0 positive solutions corresponding to the number of possible triangles given the data. It will have two positive solutions ifb sinγ <c <b, only one positive solution ifc =b sinγ, and no solution ifc <b sinγ. These different cases are also explained by theside-side-angle congruence ambiguity.

History

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Book II of Euclid'sElements, compiled c. 300 BC from material up to a century or two older, contains a geometric theorem corresponding to the law of cosines but expressed in the contemporary language of rectangle areas; Hellenistic trigonometry developed later, and sine and cosine per se first appeared centuries afterward in India.

The cases ofobtuse triangles and acute triangles (corresponding to the two cases of negative or positive cosine) are treated separately, in Propositions II.12 and II.13:^[1]

Proposition 12.
In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle.

— Euclid'sElements, translation byThomas L. Heath.^[1]

Proposition 13 contains an analogous statement for acute triangles. In his (now-lost and only preserved through fragmentary quotations) commentary,Heron of Alexandria provided proofs of theconverses of both II.12 and II.13.^[2]

Fig. 2 – Obtuse triangleABC with perpendicularBH

Using notation as in Fig. 2, Euclid's statement of proposition II.12 can be represented more concisely (though anachronistically) by the formula

$AB^{2}=CA^{2}+CB^{2}+2(CA)(CH).$

To transform this into the familiar expression for the law of cosines, substitute⁠ $AB=c$ ⁠,⁠ $CA=b$ ⁠,⁠ $CB=a$ ⁠, and $CH=a\cos(\pi -\gamma )\$ $\!{}=-a\cos \gamma$ .

Proposition II.13 was not used in Euclid's time for thesolution of triangles, but later it was used that way in the course of solving astronomical problems byal-Bīrūnī (11th century) andJohannes de Muris (14th century).^[3] Something equivalent to thespherical law of cosines was used (but not stated in general) byal-Khwārizmī (9th century),al-Battānī (9th century), andNīlakaṇṭha (15th century).^[4]

The 13th century Persian mathematicianNaṣīr al-Dīn al-Ṭūsī, in hisKitāb al-Shakl al-qattāʴ (Book on the Complete Quadrilateral, c. 1250), systematically described how to solve triangles from various combinations of given data. Given two sides and their included angle in a scalene triangle, he proposed finding the third side by dropping a perpendicular from the vertex of one of the unknown angles to the opposite base, reducing the problem to finding the legs of one right triangle from a known angle and hypotenuse using thelaw of sines and then finding the hypotenuse of another right triangle from two known sides by thePythagorean theorem.^[5]

About two centuries later, another Persian mathematician,Jamshīd al-Kāshī, who computed the most accurate trigonometric tables of his era, also described the solution of triangles from various combinations of given data in hisMiftāḥ al-ḥisāb (Key of Arithmetic, 1427), and repeated essentially al-Ṭūsī's method, now consolidated into one formula and including more explicit details, as follows:^[6]

Al-Kashi's version of the law of cosines (case whereγ is obtuse), expressed with modern algebraic notation.

Another case is when two sides and the angle between them are known and the rest are unknown. We multiply one of the sides by the sine of the [known] angle one time and by the sine of its complement the other time converted and we subtract the second result from the other side if the angle is acute and add it if the angle is obtuse. We then square the result and add to it the square of the first result. We take the square root of the sum to get the remaining side....

— Al-Kāshī'sMiftāḥ al-ḥisāb,
translation by Nuh Aydin, Lakhdar Hammoudi, and Ghada Bakbouk^[7]

Using modern algebraic notation and conventions this might be written

$c={\sqrt {(b-a\cos \gamma )^{2}+(a\sin \gamma )^{2}}}$

when⁠ $\gamma$ ⁠ is acute or

$c={\sqrt {\left(b+a\left|\cos \gamma \right|\right)^{2}+\left(a\sin \gamma \right)^{2}}}$

when⁠ $\gamma$ ⁠ is obtuse. (When⁠ $\gamma$ ⁠ is obtuse, the modern convention is that⁠ $\cos \gamma$ ⁠ is negative and $\cos(\pi -\gamma )=-\cos \gamma$ is positive; historically sines and cosines were considered to be line segments with non-negative lengths.) By squaring both sides,expanding the squared binomial, and then applying the Pythagorean trigonometric identity⁠ $\cos ^{2}\gamma +\sin ^{2}\gamma =1$ ⁠, we obtain the familiar law of cosines:

${\begin{aligned}c^{2}&=b^{2}-2ba\cos \gamma +a^{2}\cos ^{2}\gamma +a^{2}\sin ^{2}\gamma \\[5mu]&=a^{2}+b^{2}-2ab\cos \gamma .\end{aligned}}$

InFrance, the law of cosines is sometimes referred to as thethéorème d'Al-Kashi.^[8]^[9]

The same method used by al-Ṭūsī appeared in Europe as early as the 15th century, inRegiomontanus'sDe triangulis omnimodis (On Triangles of All Kinds, 1464), a comprehensive survey of plane and spherical trigonometry known at the time.^[10]

The theorem was first written using algebraic notation byFrançois Viète in the 16th century. At the beginning of the 19th century, modern algebraic notation allowed the law of cosines to be written in its current symbolic form.^[11]

Proofs

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Using the Pythagorean theorem

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Case of an obtuse angle

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Euclid proved this theorem by applying thePythagorean theorem to each of the two right triangles in Fig. 2 (AHB andCHB). Usinga to denote the line segmentCB,b to denote the line segmentAC,c to denote the line segmentAB,d to denote the line segmentCH andh for the heightBH, triangleAHB gives us $c^{2}=(b+d)^{2}+h^{2},$

and triangleCHB gives $d^{2}+h^{2}=a^{2}.$

Expanding the first equation gives $c^{2}=b^{2}+2bd+d^{2}+h^{2}.$

Substituting the second equation into this, the following can be obtained: $c^{2}=a^{2}+b^{2}+2bd.$

This is Euclid's Proposition 12 from Book 2 of theElements.^[12] To transform it into the modern form of the law of cosines, note that $d=a\cos(\pi -\gamma )=-a\cos \gamma .$

Case of an acute angle

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Euclid's proof of his Proposition 13 proceeds along the same lines as his proof of Proposition 12: he applies the Pythagorean theorem to both right triangles formed by dropping the perpendicular onto one of the sides enclosing the angleγ and uses the square of a difference to simplify.

Another proof in the acute case

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Fig. 6 – A short proof using trigonometry for the case of an acute angle

Using more trigonometry, the law of cosines can be deduced by using the Pythagorean theorem only once. In fact, by using the right triangle on the left hand side of Fig. 6 it can be shown that:

${\begin{aligned}c^{2}&=(b-a\cos \gamma )^{2}+(a\sin \gamma )^{2}\\&=b^{2}-2ab\cos \gamma +a^{2}\cos ^{2}\gamma +a^{2}\sin ^{2}\gamma \\&=b^{2}+a^{2}-2ab\cos \gamma ,\end{aligned}}$

using thetrigonometric identity⁠ $\cos ^{2}\gamma +\sin ^{2}\gamma =1$ ⁠.

This proof needs a slight modification ifb <a cos(γ). In this case, the right triangle to which the Pythagorean theorem is applied movesoutside the triangleABC. The only effect this has on the calculation is that the quantityb −a cos(γ) is replaced bya cos(γ) −b. As this quantity enters the calculation only through its square, the rest of the proof is unaffected. However, this problem only occurs whenβ is obtuse, and may be avoided by reflecting the triangle about the bisector ofγ.

Referring to Fig. 6 it is worth noting that if the angle opposite sidea isα then: $\tan \alpha ={\frac {a\sin \gamma }{b-a\cos \gamma }}.$

This is useful for direct calculation of a second angle when two sides and an included angle are given.

From three altitudes

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Fig. 5 – An acute triangle with perpendicular

Thealtitude through vertexC is a segment perpendicular to sidec. The distance from the foot of the altitude to vertexA plus the distance from the foot of the altitude to vertexB is equal to the length of sidec (see Fig. 5). Each of these distances can be written as one of the other sides multiplied by the cosine of the adjacent angle,^[13] $c=a\cos \beta +b\cos \alpha .$

(This is still true ifα orβ is obtuse, in which case the perpendicular falls outside the triangle.) Multiplying both sides byc yields $c^{2}=ac\cos \beta +bc\cos \alpha .$

The same steps work just as well when treating either of the other sides as the base of the triangle: ${\begin{aligned}a^{2}&=ac\cos \beta +ab\cos \gamma ,\\[3mu]b^{2}&=bc\cos \alpha +ab\cos \gamma .\end{aligned}}$

Taking the equation for⁠ $c^{2}$ ⁠ and subtracting the equations for⁠ $b^{2}$ ⁠ and⁠ $a^{2}$ ⁠, ${\begin{aligned}c^{2}-a^{2}-b^{2}&={\color {BlueGreen}{\cancel {\color {Black}ac\cos \beta }}}+{\color {Peach}{\cancel {\color {Black}bc\cos \alpha }}}-{\color {BlueGreen}{\cancel {\color {Black}ac\cos \beta }}}-{\color {Peach}{\cancel {\color {Black}bc\cos \alpha }}}-2ab\cos \gamma \\c^{2}&=a^{2}+b^{2}-2ab\cos \gamma .\end{aligned}}$

This proof is independent of thePythagorean theorem, insofar as it is based only on the right-triangle definition of cosine and obtains squared side lengths algebraically. Other proofs typically invoke the Pythagorean theorem explicitly, and are more geometric, treatinga cosγ as a label for the length of a certain line segment.^[13]

Unlike many proofs, this one handles the cases of obtuse and acute anglesγ in a unified fashion.

Cartesian coordinates

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Consider a triangle with sides of lengtha,b,c, whereθ is the measurement of the angle opposite the side of lengthc. This triangle can be placed on theCartesian coordinate system with sidea aligned along thex axis and angleθ placed at the origin, by plotting the components of the 3 points of the triangle as shown in Fig. 4: $A=(b\cos \theta ,b\sin \theta ),B=(a,0),{\text{ and }}C=(0,0).$

By thedistance formula,^[14]

$c={\sqrt {(a-b\cos \theta )^{2}+(0-b\sin \theta )^{2}}}.$

Squaring both sides and simplifying ${\begin{aligned}c^{2}&=(a-b\cos \theta )^{2}+(-b\sin \theta )^{2}\\&=a^{2}-2ab\cos \theta +b^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta \\&=a^{2}+b^{2}(\sin ^{2}\theta +\cos ^{2}\theta )-2ab\cos \theta \\&=a^{2}+b^{2}-2ab\cos \theta .\end{aligned}}$

An advantage of this proof is that it does not require the consideration of separate cases depending on whether the angleγ is acute, right, or obtuse. However, the cases treated separately inElements II.12–13 and later by al-Ṭūsī, al-Kāshī, and others could themselves be combined by using concepts of signed lengths and areas and a concept of signed cosine, without needing a full Cartesian coordinate system.

Using Ptolemy's theorem

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Referring to the diagram, triangleABC with sidesAB =c,BC =a andAC =b is drawn inside its circumcircle as shown. TriangleABD is constructed congruent to triangleABC withAD =BC andBD =AC. Perpendiculars fromD andC meet baseAB atE andF respectively. Then: ${\begin{aligned}&BF=AE=BC\cos {\hat {B}}=a\cos {\hat {B}}\\\Rightarrow \ &DC=EF=AB-2BF=c-2a\cos {\hat {B}}.\end{aligned}}$

Now the law of cosines is rendered by a straightforward application ofPtolemy's theorem tocyclic quadrilateralABCD: ${\begin{aligned}&AD\times BC+AB\times DC=AC\times BD\\\Rightarrow \ &a^{2}+c(c-2a\cos {\hat {B}})=b^{2}\\\Rightarrow \ &a^{2}+c^{2}-2ac\cos {\hat {B}}=b^{2}.\end{aligned}}$

Plainly if angleB isright, thenABCD is a rectangle and application of Ptolemy's theorem yields thePythagorean theorem: $a^{2}+c^{2}=b^{2}.$

By comparing areas

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Fig. 7a – Proof of the law of cosines for acute angleγ by "cutting and pasting".

Fig. 7b – Proof of the law of cosines for obtuse angleγ by "cutting and pasting".

One can also prove the law of cosines by calculatingareas. The change of sign as the angleγ becomes obtuse makes a case distinction necessary.

Recall that

a²,b², andc² are the areas of the squares with sidesa,b, andc, respectively;
ifγ is acute, thenab cosγ is the area of theparallelogram with sidesa andb forming an angle ofγ′ =⁠π/2⁠ −γ;
ifγ is obtuse, and socosγ is negative, then−ab cosγ is the area of theparallelogram with sidesa andb forming an angle ofγ′ =γ −⁠π/2⁠.

Acute case. Figure 7a shows aheptagon cut into smaller pieces (in two different ways) to yield a proof of the law of cosines. The various pieces are

in pink, the areasa²,b² on the left and the areas2ab cosγ andc² on the right;
in blue, the triangleABC, on the left and on the right;
in grey, auxiliary triangles, allcongruent toABC, an equal number (namely 2) both on the left and on the right.

The equality of areas on the left and on the right gives $a^{2}+b^{2}=c^{2}+2ab\cos \gamma .$

Obtuse case. Figure 7b cuts ahexagon in two different ways into smaller pieces, yielding a proof of the law of cosines in the case that the angleγ is obtuse. We have

in pink, the areasa²,b², and−2ab cosγ on the left andc² on the right;
in blue, the triangleABC twice, on the left, as well as on the right.

The equality of areas on the left and on the right gives $a^{2}+b^{2}-2ab\cos(\gamma )=c^{2}.$

The rigorous proof will have to include proofs that various shapes arecongruent and therefore have equal area. This will use the theory ofcongruent triangles.

Using circle geometry

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Fig. 8a – The triangleABC (pink), an auxiliary circle (light blue) and an auxiliary right triangle (yellow)

Fig. 8b – The triangleABC (pink), an auxiliary circle (light blue) and two auxiliary right triangles (yellow)

Fig. 9 – Proof of the law of cosines using the power of a point theorem.

Using thegeometry of the circle, it is possible to give a moregeometric proof than using thePythagorean theorem alone.Algebraic manipulations (in particular thebinomial theorem) are avoided.

Case of acute angleγ, wherea > 2b cosγ. Drop theperpendicular fromA ontoa =BC, creating a line segment of lengthb cosγ. Duplicate theright triangle to form theisosceles triangleACP. Construct thecircle with centerA and radiusb, and itstangenth =BH throughB. The tangenth forms a right angle with the radiusb (Euclid'sElements: Book 3, Proposition 18; or seehere), so the yellow triangle in Figure 8 is right. Apply thePythagorean theorem to obtain $c^{2}=b^{2}+h^{2}.$

Then use thetangent secant theorem (Euclid'sElements: Book 3, Proposition 36), which says that the square on the tangent through a pointB outside the circle is equal to the product of the two lines segments (fromB) created by anysecant of the circle throughB. In the present case:BH² =BC·BP, or $h^{2}=a(a-2b\cos \gamma ).$

Substituting into the previous equation gives the law of cosines: $c^{2}=b^{2}+a(a-2b\cos \gamma ).$

Note thath² is thepower of the pointB with respect to the circle. The use of the Pythagorean theorem and the tangent secant theorem can be replaced by a single application of thepower of a point theorem.

Case of acute angleγ, wherea < 2b cosγ. Drop theperpendicular fromA ontoa =BC, creating a line segment of lengthb cosγ. Duplicate theright triangle to form theisosceles triangleACP. Construct thecircle with centerA and radiusb, and achord throughB perpendicular toc =AB, half of which ish =BH. Apply thePythagorean theorem to obtain $b^{2}=c^{2}+h^{2}.$

Now use thechord theorem (Euclid'sElements: Book 3, Proposition 35), which says that if two chords intersect, the product of the two line segments obtained on one chord is equal to the product of the two line segments obtained on the other chord. In the present case:BH² =BC·BP, or $h^{2}=a(2b\cos \gamma -a).$

Substituting into the previous equation gives the law of cosines: $b^{2}=c^{2}+a(2b\cos \gamma -a).$

Note that the power of the pointB with respect to the circle has the negative value−h².

Case of obtuse angleγ. This proof uses the power of a point theorem directly, without the auxiliary triangles obtained by constructing a tangent or a chord. Construct a circle with centerB and radiusa (see Figure 9), which intersects thesecant throughA andC inC andK. Thepower of the pointA with respect to the circle is equal to bothAB² −BC² andAC·AK. Therefore, ${\begin{aligned}c^{2}-a^{2}&{}=b(b+2a\cos(\pi -\gamma ))\\&{}=b(b-2a\cos \gamma ),\end{aligned}}$

which is the law of cosines.

Using algebraic measures for line segments (allowingnegative numbers as lengths of segments) the case of obtuse angle (CK > 0) and acute angle (CK < 0) can be treated simultaneously.

Using the law of sines

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The law of cosines can be proven algebraically from thelaw of sines and a few standard trigonometric identities.^[15] To start, three angles of a triangle sum to astraight angle ( $\alpha +\beta +\gamma =\pi$ radians). Thus by the angle sum identities for sine and cosine,

${\begin{alignedat}{3}\sin \gamma &={\phantom {-}}\sin(\pi -\gamma )&&={\phantom {-}}\sin(\alpha +\beta )&&=\sin \alpha \,\cos \beta +\cos \alpha \,\sin \beta ,\\[5mu]\cos \gamma &=-\cos(\pi -\gamma )&&=-\cos(\alpha +\beta )&&=\sin \alpha \,\sin \beta -\cos \alpha \,\cos \beta .\end{alignedat}}$

Squaring the first of these identities, then substituting $\cos \alpha \,\cos \beta ={}$ $\sin \alpha \,\sin \beta -\cos \gamma$ from the second, and finally replacing $\cos ^{2}\alpha +\sin ^{2}\alpha ={}$ $\cos ^{2}\beta +\sin ^{2}\beta =1,$ thePythagorean trigonometric identity, we have:

${\begin{aligned}\sin ^{2}\gamma &=(\sin \alpha \,\cos \beta +\cos \alpha \,\sin \beta )^{2}\\[3mu]&=\sin ^{2}\alpha \,\cos ^{2}\beta +2\sin \alpha \,\sin \beta \,\cos \alpha \,\cos \beta +\cos ^{2}\alpha \,\sin ^{2}\beta \\[3mu]&=\sin ^{2}\alpha \,\cos ^{2}\beta +2\sin \alpha \,\sin \beta (\sin \alpha \,\sin \beta -\cos \gamma )+\cos ^{2}\alpha \,\sin ^{2}\beta \\[3mu]&=\sin ^{2}\alpha (\cos ^{2}\beta +\sin ^{2}\beta )+\sin ^{2}\beta (\cos ^{2}\alpha +\sin ^{2}\alpha )-2\sin \alpha \,\sin \beta \,\cos \gamma \\[3mu]&=\sin ^{2}\alpha +\sin ^{2}\beta -2\sin \alpha \,\sin \beta \,\cos \gamma .\end{aligned}}$

The law of sines holds that ${\frac {a}{\sin \alpha {\vphantom {\beta }}}}={\frac {b}{\sin \beta }}={\frac {c}{\sin \gamma {\vphantom {\beta }}}}=k,$

so to prove the law of cosines, we multiply both sides of our previous identity by⁠ $k^{2}$ ⁠:

${\begin{aligned}\sin ^{2}\gamma {\frac {c^{2}}{\sin ^{2}\gamma }}&=\sin ^{2}\alpha {\frac {a^{2}}{\sin ^{2}\alpha }}+\sin ^{2}\beta {\frac {b^{2}}{\sin ^{2}\beta }}-2\sin \alpha \,\sin \beta \,\cos \gamma {\frac {ab}{\sin \alpha \,\sin \beta {\vphantom {\sin ^{2}}}}}\\[10mu]c^{2}&=a^{2}+b^{2}-2ab\cos \gamma .\end{aligned}}$

This concludes the proof.

Using vectors

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Triangle with vector edgesa andb, separated by angleθ.

Denote

${\overrightarrow {CB}}={\vec {a}},\ {\overrightarrow {CA}}={\vec {b}},\ {\overrightarrow {AB}}={\vec {c}}$

Therefore, ${\vec {c}}={\vec {a}}-{\vec {b}}$

Taking thedot product of each side with itself: ${\begin{aligned}{\vec {c}}\cdot {\vec {c}}&=({\vec {a}}-{\vec {b}})\cdot ({\vec {a}}-{\vec {b}})\\\Vert {\vec {c}}\Vert ^{2}&=\Vert {\vec {a}}\Vert ^{2}+\Vert {\vec {b}}\Vert ^{2}-2\,{\vec {a}}\cdot {\vec {b}}\end{aligned}}$

Using the identity

${\vec {u}}\cdot {\vec {v}}=\Vert {\vec {u}}\Vert \,\Vert {\vec {v}}\Vert \cos \angle ({\vec {u}},\ {\vec {v}})$

leads to

$\Vert {\vec {c}}\Vert ^{2}=\Vert {\vec {a}}\Vert ^{2}+{\Vert {\vec {b}}\Vert }^{2}-2\,\Vert {\vec {a}}\Vert \!\;\Vert {\vec {b}}\Vert \cos \angle ({\vec {a}},\ {\vec {b}})$

The result follows.

Isosceles case

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Whena =b, i.e., when the triangle isisosceles with the two sides incident to the angleγ equal, the law of cosines simplifies significantly. Namely, becausea² +b² = 2a² = 2ab, the law of cosines becomes $\cos \gamma =1-{\frac {c^{2}}{2a^{2}}}$

or $c^{2}=2a^{2}(1-\cos \gamma ).$

Analogue for tetrahedra

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Given an arbitrarytetrahedron whose four faces have areasA,B,C, andD, withdihedral angle⁠ $\varphi _{ab}$ ⁠ between facesA andB, etc., a higher-dimensional analogue of the law of cosines is:^[16] $A^{2}=B^{2}+C^{2}+D^{2}-2\left(BC\cos \varphi _{bc}+CD\cos \varphi _{cd}+DB\cos \varphi _{db}\right).$

Version suited to small angles

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When the angle,γ, is small and the adjacent sides,a andb, are of similar length, the right hand side of the standard form of the law of cosines is subject tocatastrophic cancellation in numerical approximations. In situations where this is an important concern, a mathematically equivalent version of the law of cosines, similar to thehaversine formula, can prove useful: ${\begin{aligned}c^{2}&=(a-b)^{2}+4ab\sin ^{2}\left({\frac {\gamma }{2}}\right)\\&=(a-b)^{2}+4ab\operatorname {haversin} (\gamma ).\end{aligned}}$

In the limit of an infinitesimal angle, the law of cosines degenerates into thecircular arc length formula,c =aγ.

In non-Euclidean geometry

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Main articles:Spherical law of cosines andHyperbolic law of cosines

Spherical triangle solved by the law of cosines.

As in Euclidean geometry, one can use the law of cosines to determine the anglesA,B,C from the knowledge of the sidesa,b,c. In contrast to Euclidean geometry, the reverse is also possible in both non-Euclidean models: the anglesA,B,C determine the sidesa,b,c.

A triangle is defined by three pointsu,v, andw on the unit sphere, and the arcs ofgreat circles connecting those points. If these great circles make anglesA,B, andC with opposite sidesa,b,c then thespherical law of cosines asserts that all of the following relationships hold:

${\begin{aligned}\cos a&=\cos b\cos c+\sin b\sin c\cos A\\\cos A&=-\cos B\cos C+\sin B\sin C\cos a\\\cos a&={\frac {\cos A+\cos B\cos C}{\sin B\sin C}}.\end{aligned}}$

Inhyperbolic geometry, a pair of equations are collectively known as thehyperbolic law of cosines. The first is $\cosh a=\cosh b\cosh c-\sinh b\sinh c\cos A$

wheresinh andcosh are thehyperbolic sine and cosine, and the second is $\cos A=-\cos B\cos C+\sin B\sin C\cosh a.$

The length of the sides can be computed by:

$\cosh a={\frac {\cos A+\cos B\cos C}{\sin B\sin C}}.$

Polyhedra

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The law of cosines can be generalized to allpolyhedra by considering any polyhedron with vector sides and invoking thedivergence theorem.^[17]

Notes

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^Given sides⁠ $b {\displaystyle b}$ ⁠,⁠ $c {\displaystyle c}$ ⁠ and angle⁠ $\gamma$ ⁠,⁠ $\sin \beta$ ⁠ can be found using the law of sines, leaving up to two possibilities for angle⁠ $\beta$ ⁠. Either choice determines⁠ $\alpha$ ⁠ because the three interior angles sum to a straight angle. Finally⁠ $a {\displaystyle a}$ ⁠ can be found from⁠ $c {\displaystyle c}$ ⁠,⁠ $\gamma$ ⁠, and⁠ $\alpha$ ⁠ by another application of the law of sines.

References

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^^a ^bEuclid.Thomas L. Heath (ed.)."Elements". Translated by Thomas L. Heath. Retrieved24 January 2023.
^Heath, Thomas (1956) [1908]."Introduction".The Thirteen Books of Euclid's Elements (2nd ed.).
^Kennedy, E.S.; Muruwwa, Ahmad (1958). "Bīrūnī on the Solar Equation".Journal of Near Eastern Studies.17 (2):112–121.doi:10.1086/371451.JSTOR 542617.
Johannes de Muris credits an anonymous author for the relevant section of his workDe Arte Mesurandi. SeeVan Brummelen, Glen (2009).The Mathematics of the Heavens and the Earth. Princeton University Press. pp. 240–241.
^Van Brummelen, Glen (2012).Heavenly mathematics: The forgotten art of spherical trigonometry. Princeton University Press. p. 98.
^Naṣīr al-Dīn al-Ṭūsī (1891)."Ch. 3.2: Sur la manière de calculer les côtés et les angles d'un triangle les uns par les autres".Traité du quadrilatère attribué a Nassiruddinel-Toussy (in French). Translated byCaratheodory, Alexandre Pacha. Typographie et Lithographie Osmanié. p. 69.On donne deux côtés et un angle. [...] Que si l'angle donné est compris entre les deux côtés donnés, comme l'angle A est compris entre les deux côtés AB AC, abaissez de B sur AC la perpendiculaire BE. Vous aurez ainsi le triangle rectangle [BEA] dont nous connaissons le côté AB et l'angle A; on en tirera BE, EA, et l'on retombera ainsi dans un des cas précédents; c. à. d. dans le cas où BE, CE sont connus; on connaîtra dès lors BC et l'angle C, comme nous l'avons expliqué [Given [...] the angle A is included between the two sides AB AC, drop from B to AC the perpendicular BE. You will thus have the right triangle [BEA] of which we know the side AB and the angle A; in that triangle compute BE, EA, and the problem is reduced to one of the preceding cases; that is, to the case where BE, CE are known; we will thus know BC and the angle C, as we have explained.]
^Azarian, Mohammad K. (2000)."Meftab Al-Hesab: A Summary"(PDF).Missouri Journal of Mathematical Sciences.12 (2):75–95.doi:10.35834/2000/1202075.
^Aydin, Nuh; Hammoudi, Lakhdar; Bakbouk, Ghada (2020).Al-Kashi's Miftah al-Hisab, Volume II: Geometry. Birkhäuser. p. 31.doi:10.1007/978-3-030-61330-3.ISBN 978-3-030-61329-7.
^Pickover, Clifford A. (2009).The Math Book: From Pythagoras to the 57th Dimension. Sterling Publishing Company, Inc. p. 106.ISBN 9781402757969.
^Programme de mathématiques de première générale (in French). Ministère de l'Éducation nationale et de la Jeunesse. 2022. pp. 11, 12.
^Hughes, Barnabas, ed. (1967).Regiomontanus on Triangles. Madison: University of Wisconsin Press. English translation ofDe triangulis omnimodis by Johann Müller, otherwise known as Regiomontanus, with facsimile of the 1533 Latin edition.Theorem I.49, pp. 98–101.
^For example inCarnot, Lazare (1803).Géométrie de position. J.B.M Duprat. p. 202.
^Java applet version by Prof.D E Joyce of Clark University.
^^a ^bAlexander Bogomolny credits this proof to teacher John Molokach (2011), but it may be older.Bogomolny, Alexander."The Law of Cosines (Independent of the Pythagorean Theorem)".Cut the Knot. Retrieved2024-01-09.
^Wylie, Clarence Raymond (1955).Plane Trigonometry. McGraw-Hill. §9.1 The Law of Cosines, pp. 195–198.LCCN 54-11278.
^Burton, L. J. (1949). "The Laws of Sines and Cosines".The American Mathematical Monthly.56 (8):550–551.doi:10.1080/00029890.1949.11999439.JSTOR 2305533.
^Casey, John (1889).A Treatise on Spherical Trigonometry: And Its Application to Geodesy and Astronomy with Numerous Examples. London: Longmans, Green, & Company. p. 133.
^Collins, L; Osler, T (2011). "Law of Cosines Generalised for any Polygon and any Polyhedron".The Mathematical Gazette.95 (533):240–243.doi:10.1017/S0025557200002953.JSTOR 23248682.

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