# Trigonometric Identities#

There are a large number of trigonometric identities that one can memorize, but I tend to remember only a few bare essentials. Specifically, I remember the following angle-addition identities: $\( \begin{aligned} \sin(x \pm y) &= \sin x \cos y \pm \sin y \cos x \\ \cos(x \pm y) &= \cos x \cos y \mp \sin x \sin y \end{aligned} \)\( Note that in fact you need to memorize only the identities with the \)+\( sign, because the others can be deduced by changing \)y \rightarrow -y\( and remembering that the cosine is an even function and the sine is an odd function, \)\( \begin{aligned} \cos(-x) &= \cos(x) \\ \sin(-x) &= - \sin(x) \end{aligned} \)\( So in fact I only "memorize" \)\( \begin{aligned} \sin(x + y) &= \sin x \cos y + \sin y \cos x \\ \cos(x + y) &= \cos x \cos y - \sin x \sin y \end{aligned} \)$

From these I can deduce many other identities very easily. For example, by choosing \(x=y\) in the angle-addition identities, I can deduce the double-angle identities and the Pythagorean identity: $\( \begin{aligned} \sin(2x) &= 2 \sin x \cos x \\ \cos(2x) &= \cos^2 x - \sin^2 x \\ \cos(0) &= 1 = \cos^2 x + \sin^x \end{aligned} \)\( Adding the second two equations gives \)\( \cos 2x + 1 = 2 \cos^2 x \)\( and subtracting the second two equations gives \)\( 1 - \cos 2x = 2 \sin^2 x \)\( Alternatively, going back to the angle-addition identities, you can deduce (by adding and subtracting each equation from itself) the product-to-sum identities: \)\( \begin{aligned} 2 \sin x \cos y &= \sin(x+y) + \sin(x-y) \\ 2 \sin x \sin y &= \cos(x-y) - \cos(x+y) \\ 2 \cos x \cos y &= \cos(x-y) + \cos(x+y) \end{aligned} \)\( The double-angle formulas could also be deduced by setting \)x = y\( in these equations. One can also deduce the sum-to-product identities: \)\( \begin{aligned} \sin x + \sin y &= 2\sin \left(\tfrac{a + b}{2}\right)\cos \left(\tfrac{a - b}{2}\right) \\ \cos x + \cos y &= 2\cos \left(\tfrac{a + b}{2}\right)\cos \left(\tfrac{a - b}{2}\right) \\ \cos x - \cos y &= -2\sin \left(\tfrac{a + b}{2}\right)\sin \left(\tfrac{a - b}{2}\right) \end{aligned} \)$