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(2006-11-20)   The Trigonometric Circle Diagram
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 for the reciprocal of the tangent.

The names of the reciprocals of the sine and cosine functions are deprecated.  If you must know, the secant (sec)  is the reciprocal of the cosine and the cosecant (csc or cosec) is the reciprocal of the sine function.  For the record:

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

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

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

(2002-01-29)   Trigonometry in a Nutshell
What are the basic laws used with trigonometric functions to 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 r of the circumscribed circle (see figure at right). The angular data is usually (but not always) expressed directly in terms of the inside angles between sides  (the angles a, b, g, which add up to p).

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

The Sine Law may well be the most useful of the three.  It delivers directly any missing quantity in all basic cases not covered by the next paragraph...

If the 3 sides are given, the Law of Cosines gives you any angle you may want (via its cosine). The Law of Cosines will also give you the missing side (c) when a, b and g are given  (the "SAS case", in high-school parlance).  Having the three sides, you could then obtain either of the missing angles by using the Law of Cosines again. However, it's more elegant and more direct to compute the missing angles with the Law of Tangents (especially, if you do not care about the value of the missing side c): Since you know a/2+b/2 (it's equal to p/2-g/2), the Law of Tangents gives you the (tangent of the) angle a/2-b/2. and the missing angles are simply the sum and the difference of a/2+b/2 and a/2-b/2.

Proofs : The Law 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  r  forming an angle 2a.  The length of the base  (a)  is twice the side opposite to an angle  a  in a right triangle of hypotenuse  r.

The above proof was published in 1342 by the  Provençal  rabbi  Levi ben Gerson (1288-1344)  who is variously known as "RaLBaG", Levi ben Gershon, Léon de Bagnols, Magister Leo Judaeus, or  Gersonides  [son of Gershon ben Solomon d'Arles].
 
The  sine law  is often attributed to Abu Nasr Mansur (970-1036) who either discovered it himself or got it from Abu'l-Wafa (940-998).  Abu Nasr Mansur  was the mathematical mentor and partner of Al-Biruni (973-1048).  Historians now agree that  Nasir al-Tusi (1201-1274) was mistaken when he attributed that law to the astronomer Al-Khujandi (940-1000) who never made such a claim himself.

The Law of Cosines is best proved (or memorized) in the modern context of vector algebra:  Consider two vectors U (from A to C) and V (from A to B), the vector  U-V goes from B to C and its square (U-V)²  is

|U|² - 2 U.V + |V|²

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

More traditionally, the law of cosines can be construed to be equivalent to  Heron's formula  for the surface area of a triangle,  which appears in Book I of the  Metrica  by  Heron of Alexandria (c. AD 10-75).  However, it would be quite a stretch to make it go back this far, since Heron didn't know what a cosine was...

The Law of Tangents was first stated around 1580 by François Viète (Viette, or Vieta).  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 a and sin b, using the Law of Sines.  The result is then immediately obtained from the following identity for the denominator and from its counterpart for the numerator (replace b by -b).

sin a + sin b   =   2 sin(a/2 + b/2) cos(a/2 - b/2)  

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

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

Menelaus of Alexandria,  who originated the subject around AD 100,  needlessly required that each side should be less than a semicircle.

Each such figure divides the surface of the sphere into two parts, whose areas add up to 4pR² .  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 called spherical trigonometry and is about as ancient as the simpler planar trigonometry summarized above.

The internal angles of a spherical triangle always add up to more than a flat angle (of p radians).  Expressed in radians, the difference (denoted e, with 0< e <4p) is usually called the spherical excess, a term coined around 1626 by the French-born Dutch mathematician Albert Girard (1595-1632), who showed that the surface area of a spherical triangle is simply equal to:

e R ²   =   ( (a + b + g) - p ) R ²

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

sin a	=	sin b	=	sin c
sin a		sin b		sin g

In this, a, b and c are the angular "lengths" of the sides (as seen from the sphere's center); they are the curvilinear distances along the great arcs, using R as a unit.  The spherical excess may also be expressed in terms of these quantities and the semiperimeter  s = ½ (a+b+c), using the spherical equivalent of Hero's formula:

L'Huilier's Theorem

[tg(e / 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 mathematician Simon L'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 are symmetrical with respect to u and v, may be expressed in 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 or X²+(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 the following 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 Watts of Hickory, NC. 2001-03-11)
Show that   | sin (x + iy) | ²   =   sin ²(x) + sinh ²(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 of the 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 ² (x)   =   1/2 - cos(2x)/2
• sinh ² (y)   =   (e ^2y + e^-2y )/4 - 1/2

Therefore, the entire expression is indeed sin²(x) + sh²(y), as advertised.

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

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

If p < q, then     Ö[p/q]	=   sin arctan Ö[p/(q-p)]
If p > q, then     Ö[p/q]	=   tan arccos Ö[q/(p+q)] =   tan arccos sin arctan Ö[q/p] =   tan arccos sin arctan sin arctan Ö[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 rationals can 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 of 25/64,  which is the  sin arctan  of the square root of 25/39, itself the  sin arctan  of the square root of 25/14, itself the  tan arccos sin arctan sin arctan  of the square root of 14/11, itself the  tan arccos sin arctan sin arctan  of the square root of 11/3, itself the  tan arccos sin arctan sin arctan  of the square root of 3/8, itself the  sin arctan  of the square root of 3/5, itself the  sin arctan  of the square root of 3/2, itself the  tan arccos sin arctan sin arctan  of the square root of 2, itself the  tan arccos sin arctan sin arctan  of (the square 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 from inverse trigonometric functions, except for the initial zero angle  (either 0 degrees or 0 radians).