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

Does anyone know of a closed formula for $\displaystyle \cos\left(\sum_{n=0}^m a_n\right)$? I've seen formulas for $\displaystyle \cos\left(\sum_{n=0}^\infty a_n\right)$ and $ \displaystyle \tan\left(\sum_{n=0}^m a_n\right)$, 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.

\begin{align} & \cos \left( \sum_{n=0}^m a_n \right) \\[8pt] = {} & \operatorname{Re} \left[ \exp \left(i\sum_{n=0}^m a_n \right) \right] \\[8pt] = {} & \frac{1}{2} \left[\exp\left(i\sum_{n=0}^m a_n \right) + \exp\left(-i\sum_{n=0}^m a_n\right) \right] \\[8pt] = {} & \frac{1}{2} \left[ \prod_{n=0}^m \exp(ia_n) + \prod_{n=0}^m \exp(-ia_n) \right] \\[8pt] = {} & \frac{1}{2} \left[ \prod_{n=0}^m \left[\cos(a_n) + i\sin(a_n) \right] \right] + \prod_{n=0}^m \left[\cos(a_n)-i\sin(a_n)\right] \end{align}

which amounts to finding a closed-form expression for

$$(A_0+B_0)(A_1+B_1)(A_2+B_2)\cdots(A_m+B_m)$$

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:

\begin{align} & \cos \left( \sum_{n=0}^m a_n\right) \\[8pt] = {} & \cos\left(a_m +\sum_{n=0}^{m-1} a_n \right) \\[8pt] = {} & \cos(a_m) \cos\left(\sum_{n=0}^{m-1} a_n \right) - \sin(a_m) \sin\left(\sum_{n=0}^{m-1} a_n\right) \end{align}

Jackson