How can we motivate the derivations and proofs of trigonometric identities? Jacques Hadamard said "The shortest path between two truths in the real domain sometimes passes through the complex domain", and this is certainly true when it comes to verifying trigonometric identities such as A + B A - B 2 sin ----- cos ----- = sin A + sin B 2 2 Going from the left side of the equation to the right is just a matter of multiplying through using the exponential forms of sin() and cos() as follows _ _ _ _ | i(A+B)/2 -i(A+B)/2 | | i(A-B)/2 -i(A-B)/2 | | e - e | | e + e | 2 | ------------------------ | | ------------------------ | |_ 2i _| |_ 2 _| iA -iA iB -iB e - e e - e = ----------- + ---------- 2i 2i This is perfectly straightforward. However, going from the right side to the left side is essentially a task of factorization. In general the two tasks have different orders of difficulty. Asking for a "converted" version of sin(x)+sin(y) is analagous to asking for a factored version of, say, 2258745004684033. In contrast, asking for a proof of the identity sin(x)+sin(y) = 2sin((x+y)/2)cos((x-y)/2) is analagous to asking for a proof of the identity 2258745004684033 = (27439297)(82317889) A proof (or disproof) of a given proposition is generally easier than constructing the proposition in the first place. This is similar to integral formulas. Given the integral of a function it's usually easy to differentiate and find the function, but finding the integral from the function is often much more difficult. In this sense, differentiation is to multiplication as integration is to factoring. Very often in situations like this we take the approach of just differentiating every function we can think of, and write down the derivatives in a table along with the function. Then to find the integral of a function we just look for that function in the derivative column of our table. A similar approach was used to construct the tables of standard trigonometric identities. This approach have very wide applicability. I think it was Jacobi who said "Always invert!", and this is certainly how he made most of Legendre's work on eliptic integrals obsolete. (This same approach works outside of mathematics too.) Combining Hadamard and Jacobi's comments, we can say that, in general, two very powerful rules are for producing useful results quickly are: (1) Use complex numbers, and (2) always invert. Another general and highly efficatious approach to functional problems is to express them in terms of power series. The realization that a large class of functions can be represented as power series was one of the most significant turning points in the development of modern mathematics. To show how this can be applied to the derivation of trigonometric identities, suppose we want to find an expression for sin(a+b) in terms of the sines and cosines of the individual numbers a and b. For a general power series f(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + ... we have the derivatives f'(x) = c_1 + 2c_2 x + 3c_3 x^2 + ... f''(x) = 2c_2 + 6c_3 x + ... f'''(x) = 6c_3 + ... and so on. It's clear that the nth derivative at x=0 is simply (n!)c_n, and so the nth coefficient c_n equals the nth derivative divided by n!. Now recall that the derivative of the sine is the cosine, and the derivative of the cosine is the negative sine, so we can expand the function f(x) = sin(a+x) around the point x=0 as follows cos(a) sin(a) cos(a) sin(a+x) = sin(a) + ------ x - ------ x^2 - ------ x^3 + ... 1! 2! 3! By the same reasoning we can write down the power series for cos(a+x) expanded about the point a: sin(a) cos(a) cos(a) cos(a+x) = cos(a) - ------ x - ------ x^2 + ------ x^3 - ... 1! 2! 3! Not surprisingly, these expressions reduce to the familiar power series for sin(x) and cos(x) if we set a=0. Now, suppose we collect terms in the series for sin(a+x) according to whether they contain a sine or a cosine. This gives _ _ | x^2 x^4 x^6 | sin(a+x) = sin(a)| 1 - --- + --- - --- + ... | |_ 2! 4! 6! _| _ _ | x^3 x^5 x^7 | + cos(a)| x - --- + --- - --- + ... | |_ 3! 5! 7! _| Of course, we immediately recognize the quantities inside the brackets as the cosine and sine of x, so we have the trigonometric identity sin(a+b) = sin(a)cos(b) + cos(a)sin(b) Similarly if we separate the terms of the series for cos(a+x) we arrive at _ _ | x^2 x^4 x^6 | cos(a+x) = cos(a)| 1 - --- + --- - --- + ... | |_ 2! 4! 6! _| _ _ | x^3 x^5 x^7 | - sin(a)| x - --- + --- - --- + ... | |_ 3! 5! 7! _| which gives the identity cos(a+b) = cos(a)cos(b) - sin(a)sin(b) This nicely illustrates how it is often possible to deduce closed- form identities from consideration of the infinite series expansions of the functions involved.

Return to MathPages Main Menu