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(2006-11-20)  
Memorize 3 basic trigonometric functions (out of 6)

Modern usage retains mostly the three trigonometric functions depicted at right: sine (sin) cosine (cos)  and tangent (tan or tg). The cotangent function  (cot or cotg)  may be used forthe reciprocal of the tangent.

The names of the reciprocals of the sine and cosine functions aredeprecated. If you must know, thesecant (sec)  is the reciprocal of thecosine and thecosecant (csc or cosec) is the reciprocal of thesine function.  For the record:

tan	=	sin/ cos		sec	=	1/ cos
cot	=	cos/ sin		csc	=	1/ sin

Each of those 6 trigonometric functions is the ratio of two sides in a right triangle whereone of the acute angles is specified.  However, it's much better to consider theso-called trigometric circle  of unit radius depicted above,which generalizes those functions smoothly to obtuse and/or negative values of the angle  (e.g., the cosine of an obtuse angle is negative). 

Just memorize the above pictorial definitions ofthe 3 main trigonometric functions and ignore the rest (unless you expect a quizz at school, that is).

(2002-01-29)  
What are the basic laws used with trigonometric functionsto obtain all the elements of a triangle when only some of them are known?

A planar triangle is determined by 3 independent quantities.These could be the 3 sides (a,b,c,each of which is less than or equal to the sum of the other two),1 angle and 2 sides, or 1 side and 2 angles.Instead of a side, you may be given some other length related to the triangle; for example,the radius of the circumscribed circle(see figure at right).The angular data is usually (but not always) expressed directly in terms of theinside angles between sides  (the angles ,which add up to ).

The Law of Sines :	2   =   a   =   b   =   c sin sin sin
The Law of Cosines :	c2   =  a2 +b2- 2 ab cos
The Law of Tangents :	a -b   =   tg ()  a +b tg ()

TheSine Law may well be the most useful of the three. It delivers directly any missing quantity in all basic cases not coveredby the next paragraph...

If the 3 sides are given, theLaw of Cosines gives you any angleyou may want (via itscosine).TheLaw of Cosines will also give you the missing side (c)whena,b and are given (the "SAS case", in high-school parlance). Having the three sides, you could then obtain either of the missing angles by usingtheLaw of Cosines again.However, it's more elegant and more direct to computethe missing angles with theLaw of Tangents(especially, if you do not care about the value of the missing sidec):Since you know (it's equal to ),theLaw of Tangents gives you the (tangent of the) angle.and the missing angles are simply the sum and thedifference of and .

Proofs : TheLaw of Sines can be proved by remarking that,if O is the center of the circumscribed circle,one may consider an isoceles triangle like OBC which has two sides of length   forming an angle 2. The length of the base  (a)  is twice the side opposite to an angle  in a right triangle of hypotenuse  .

TheLaw of Cosines is best proved (or memorized) in the modern context ofvector algebra: Consider two vectorsU (from A to C) andV (from A to B),the vector U-Vgoes from B to C and its square (U-V)²  is

|U|² - 2U.V + |V|²

The scalar product ("dot product")U.V is equal to thecosine of the angleformed by the two vectors multiplied by the product of their magnitudes.  

TheLaw of Tangents was first stated around 1580 byFrançoisViète (Viette, orVieta). It's the least popular of the three,and is not always found in textbooks...  To prove it, consider the following ratio:

(a -b)/(a +b)

Express it in terms of sin  and sin ,using theLaw of Sines. The result is then immediately obtained from the following identity for the denominator andfrom its counterpart for the numerator(replace by ).

sin + sin   =   2sin() cos() 

(2003-06-20)    
What are some basic relations applicable to spherical triangles?

A spherical triangle is a figure on the surface ofa sphere of radius R,featuring three sides which are arcs ofgreat circles(a "great circle" is the intersection of the sphere with a plane containing thesphere's center).

Each such figure divides the surface of the sphere into two parts,whose areas add up to4R. Unless otherwise specified, the smaller part is usually considered the "inside"of the triangle, but this need not always be so...

The study of spherical triangles is often calledspherical trigonometryand is about as ancient as the simplerplanar trigonometry summarizedabove.

The internal angles of a spherical triangle always add up tomore than aflat angle (of radians). Expressed in radians, the difference(denoted , with0<  <4)is usually called thespherical excess, a term coined around 1626by the French-born Dutch mathematicianAlbertGirard (1595-1632),who showed that the surface area of a spherical triangle issimply equal to:

R ²   =  ( ) R ²

There are some striking similarities between the two kinds of trigonometries,including the sphericalLaw of Sines ofAbu'l-Wafa (940-998):

sina	=	sinb	=	sinc
sin		sin		sin

In this,a,b andc are theangular "lengths" of thesides (as seen from the sphere's center); they are thecurvilinear distances along the great arcs, using R as a unit. Thespherical excess may also be expressed in terms of these quantitiesand thesemiperimeter s = ½ (a+b+c),using the spherical equivalent ofHero's formula:

L'Huilier's Theorem

[tg( / 4)]2   =  tg[s / 2]  tg[(s-a) / 2]  tg[(s-b) / 2] tg[(s-c) / 2]

This beautiful formula is named after the Swiss mathematicianSimonL'Huilier (1750-1840) who was once a teacher of Charles-François Sturm in Geneva. (His last name is sometimes also spelled "L'Huillier" or "Lhuilier".)

(V. R. of India.2000-10-16)
Let   cos A + cos B = 2p   and   sin A + sin B = 2q .
Prove that     tan A/2 + tan B/2   =   2q / (p² + q² + p)

Let u be tan(A/2) and v be tan(B/2).  We have:

cos(A) = (1-u²)/(1+u²)
sin(A) = 2u/(1+u²)

Similar relations hold for B and v.  Therefore:

2p   =   (1-u²) / (1+u²)  +  (1-v²) / (1+v²)
2q   =   2u / (1+u²)  +  2v / (1+v²)

Expressions such as these, which aresymmetrical with respect to u and v, may be expressedin terms of the sum X=u+v and the product Y=uv.For example u²+v² is X²-2Y and(1+u²)(1+v²) is 1+X²-2Y+Y² orX²+(1-Y)².The above two relations thus become:

2p (X²+(1-Y)²)   =   (1-u²)(1+v²) + (1-v²)(1+u²)
=   2-2Y²   =   2(1-Y)(1+Y)
 
and    2q (X²+(1-Y)²)   =   (2u)(1+v²)+(2v)(1+u²)
=   2X + 2XY   =   2X (1+Y)

Adding or subtracting these two after multiplying each by either (1-Y) or X(and removing the nonzero factor X²+(1-Y)² which turns up)greatly simplifies this system of equations, which boils down to a linear system:

pX   =   q(1-Y)     and     p (1-Y)+ q X   =   (1+Y)

This may be rewritten

p X  +  q Y   =   q     and     q X  -  (p+1) Y   =   (1-p)

Solving for  X  gives   X [p(p+1)+q²]  =  q(p+1)+q(1-p)  =  2q, which is the desired relation:

tan(A/2) + tan(B/2)   =   X   =   2q / (p² + q² + p)

Also, we may as well solve for  Y  to obtain another interesting relation:

tan(A/2) tan(B/2)   =   Y   =  (p² + q² - p) / (p² + q²+p)

Those two results are equivalent to the statement that the two roots of thefollowing quadratic equation in  t  are  tan(A/2) and  tan(B/2) :

( p ² + q ² + p )  t ²    2 q t   +  ( p ² + q ² p )    =   0

brentw (Brent Wattsof Hickory, NC.2001-03-11)
Show that   | sin (x + iy) |   =  sin(x) + sinh(y)

The relation exp(x+iy) = exp(x)(cos(y)+ i sin(y))may be turned into a definition of the cosine and sine function(x and y need not be real in this).

In particular, exp(iz) = cos(z)+ i sin(z)  so,  sin(z) = (exp(iz)-exp(-iz))/2i.  Therefore:

sin(x+iy)  =  [e^ix-y - e^-ix+y] / 2i
=  [e^-y(cos(x)+ i sin(x)) - exp^y(cos(x)- i sin(x))]/2i
=  [cos(x)(e^-y - e^y) + i sin(x)(e^-y + e^y)] / 2i

So far, we did not assume that x and y were real, now we do: |z| ²  is the sum of the squares ofthe real and imaginary parts of z. When z is the last of the above expressions, this translates into

| sin(x+iy) | ²  =  [cos²(x)(e^-y - e^y)²+ sin²(x)(e^-y + e^y)²] / 4
=  [(e^-2y + e^2y) - 2 cos(2x) ] / 4

This expression can be idenfified with the given one by noticing that:

•   sin(x)   =   1/2 - cos(2x)/2
• sinh(y)   =  (e + e)/4 - 1/2

Therefore, the entire expression is indeedsin(x) + sh(y),as advertised.

FlyingHellfish(2003-07-28)
In a broken calculator, only the 6 functions shownat right are available. Can any positive rational number be obtained from an initial 0?

Actually, [p/q]can be obtained for any positive integers p and q, since:

If p < q, then    [p/q]	=   sin arctan [p/(q-p)]
If p > q, then    [p/q]	=   tan arccos [q/(p+q)] =   tan arccos sin arctan [q/p] =   tan arccos sin arctan sin arctan [q/(p-q)]

Using whichever relation is relevant, this reduces any case to a simpler one,until we're faced with  p = q, which we solve by pushing cos  once. 

There's (almost) no need to say that the above shows that all positive rationalscan be so obtained, since each is the square root of its square.

For example, if we wish to obtain 5/8, we observe that it's the square root of25/64,  which is the sin arctan  of thesquare root of 25/39, itself the sin arctan  of thesquare root of 25/14, itself the tan arccos sin arctan sin arctan  of thesquare root of 14/11, itself the tan arccos sin arctan sin arctan  of thesquare root of 11/3, itself the tan arccos sin arctan sin arctan  of thesquare root of 3/8, itself the sin arctan  of thesquare root of 3/5, itself the sin arctan  of thesquare root of 3/2, itself the tan arccos sin arctan sin arctan  of thesquare root of 2, itself the tan arccos sin arctan sin arctan  of (thesquare root of)  1,which is, of course, the cos  of 0... Only  39  keys to press on that broken calculator.

It's irrelevant whether the calculator works in degrees or in radians,since the only angles we use are obtained frominverse trigonometric functions, except for the initial zero angle (either 0 degrees or 0 radians).