Does anyone know of a closed formula for $cos(\displaystyle\sum_{n=0}^m a_{n})$? I've seen formulas for $cos(\displaystyle\sum_{n=0}^\infty a_{n})$ and $tan(\displaystyle\sum_{n=0}^m a_{n})$, but the former remains elusive. I can only think of two ways to approach the problem, either by taking the real part of complex exponentials or defining a recurrence relation, viz.
$cos(\displaystyle\sum_{n=0}^m a_{n})$
$=Re[exp(i\displaystyle\sum_{n=0}^m a_{n})]$
$=\frac{1}{2}[exp(i\displaystyle\sum_{n=0}^m a_{n})+exp(-i\displaystyle\sum_{n=0}^m a_{n})]$
$=\frac{1}{2}[\displaystyle\prod_{n=0}^mexp(ia_{n})+\displaystyle\prod_{n=0}^mexp(-ia_{n})]$
$=\frac{1}{2}[\displaystyle\prod_{n=0}^m[cos(a_{n})+isin(a_{n})])+\displaystyle\prod_{n=0}^m[cos(a_{n})-isin(a_{n})]]$
which amounts to finding a closed-form expression for
$(A_{0}+B_{0})(A_{1}+B_{1})(A_{2}+B_{2})…(A_{m}+B_{m})$
similar to finding binomial coefficients, albeit more general. I know there should be $2^m$ unique terms arising from choosing either $A_{0}$ or $B_{0}$, then choosing either $A_{1}$ or $B_{1}$, etc. until you've chosen every combination. The recurrence relation would go as follows:
$cos(\displaystyle\sum_{n=0}^m a_{n}) = cos(a_{m}+\displaystyle\sum_{n=0}^{m-1} a_{n}) = cos(a_{m})cos(\displaystyle\sum_{n=0}^{m-1} a_{n})-sin(a_{m})sin(\displaystyle\sum_{n=0}^{m-1} a_{n})$
Jackson
Best Answer
Take the formula for $\cos\left(\sum_{n=0}^\infty a_n\right)$ and let all but finitely many terms be 0. Notice that you have a sum of products of sines and cosines. Each product has only finitely many sine factors. If one of those is the sine of 0, then the whole term vanishes. But if none is the sine of 0, then you have the product of finitely many cosines of nonzero numbers and infinitely factors each of which is $\cos 0 = 1$. In effect, those $\cos 0$ factors also vanish. There you have it.
....also: Look at 19th- and early 20th-century books on trigonometry. Lot's of stuff is there that you won't find in more recent books on that topic.