Cosine of a Partial Sum

Question

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

Answer 1

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.

Answer 2

Michael, that looks like the same explanation you have on Wikipedia, which I have seen, as well as your post on this site about a related topic. While it's a correct formula for $cos(\displaystyle\sum_{n=0}^m a_{n})$, as I said in a previous comment, I am looking for a much more elementary formula that doesn't ask one to "take every permutation of a subset of natural numbers of order k" in the index without an explicit formula. If you'll bear with the indices, I think I've figured it out to my satisfaction.

In order to find

$(A_{0}+B_{0})(A_{1}+B_{1})...(A_{m}+B_{m})$

we'll need to relabel A and B and instead use $\alpha_{0}$ in place of $A$ and $\alpha_{1}$ in place of $B$. Rewriting, we need to find:

$(\alpha_{00}+\alpha_{10})(\alpha_{01}+\alpha_{11})...(\alpha_{0m}+\alpha_{1m})$

Since each term will be formed by choosing one term from each binomial, every term will be of the form $\displaystyle\prod_{k=0}^m\alpha_{j_{k}k}$ where $j_{k} \in ${$0,1$}

Thus, we may write $\displaystyle\prod_{n=0}^m (\alpha_{0n}+\alpha_{1n})$ as an 'm-sum', viz.

$\displaystyle\prod_{n=0}^m (\alpha_{0n}+\alpha_{1n}) = \displaystyle\sum_{j_{0}=0}^1\displaystyle\sum_{j_{1}=0}^1\displaystyle\sum_{j_{2}=0}^1...\displaystyle\sum_{j_{m}=0}^1(\displaystyle\prod_{k=0}^m\alpha_{j_{k}k})$

The product basically sets up the term, and the sums take you through every permutation.

As an example, in the case where m=1:

$(A_{0}+B_{0})(A_{1}+B_{1})$ $= (\alpha_{00}+\alpha_{10})(\alpha_{01}+\alpha_{11})$ $= \displaystyle\prod_{n=0}^1 (\alpha_{0n}+\alpha_{1n})$ $= \displaystyle\sum_{i_{0}=0}^1\displaystyle\sum_{i_{1}=0}^1(\displaystyle\prod_{k=0}^1\alpha_{i_{k}k})$ $= \displaystyle\sum_{i_{0}=0}^1\displaystyle\sum_{i_{1}=0}^1 \alpha_{i_{0}0}\alpha_{i_{1}1}$ $= \alpha_{00}\alpha_{01}+\alpha_{00}\alpha_{11}+\alpha_{10}\alpha_{01}+\alpha_{10}\alpha_{11}$ $= A_{0}A_{1}+A_{0}B_{1}+B_{0}A_{1}+B_{0}B_{1}$

as promised.

Define $\beta_{0n}:=cos(a_{n})$ and $\beta_{1n}:=isin(a_{n})$. Using the result from above, we see that:

$\displaystyle\prod_{n=0}^m[cos(a_{n})+isin(a_{n})]=$ $\displaystyle\sum_{j_{0}=0}^1\displaystyle\sum_{j_{1}=0}^1\displaystyle\sum_{j_{2}=0}^1...\displaystyle\sum_{j_{m}=0}^1(\displaystyle\prod_{k=0}^m\beta_{j_{k}k})$

Similarly, define $\gamma_{0n}:=cos(a_{n})$ and $\gamma_{1n}:=-isin(a_{n})$, then:

$\displaystyle\prod_{n=0}^m[cos(a_{n})-isin(a_{n})]=$ $\displaystyle\sum_{j_{0}=0}^1\displaystyle\sum_{j_{1}=0}^1\displaystyle\sum_{j_{2}=0}^1...\displaystyle\sum_{j_{m}=0}^1(\displaystyle\prod_{k=0}^m\gamma_{j_{k}k})$

Thus,

$cos(\displaystyle\sum_{n=0}^ma_{n})=\frac{1}{2}[\displaystyle\sum_{j_{0}=0}^1\displaystyle\sum_{j_{1}=0}^1\displaystyle\sum_{j_{2}=0}^1...\displaystyle\sum_{j_{m}=0}^1(\displaystyle\prod_{k=0}^m\beta_{j_{k}k})+\displaystyle\sum_{j_{0}=0}^1\displaystyle\sum_{j_{1}=0}^1\displaystyle\sum_{j_{2}=0}^1...\displaystyle\sum_{j_{m}=0}^1(\displaystyle\prod_{k=0}^m\gamma_{j_{k}k})]$

Apologizes for the explosion of indices.