MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


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:

$cos(\displaystyle\sum_{n=0}^m a_{n}) = cos(a_{m}+\displaystyle\sum_{n=0}^{m-1} a_{n}) = cos(a_{m})cos(\displaystyle\sum_{n=0}^{m-1} a_{n})-sin(a_{m})sin(\displaystyle\sum_{n=0}^{m-1} a_{n})$


share|cite|improve this question
If you have an expression for cos of the sum to infinity, why not just let $a_j=0$ for all $j>m$? – Zen Harper May 16 '11 at 1:23
Well the only formula I've seen for $cos(\displaystyle\sum_{n=0}^\infty a_{n})$ is $\sum_{\text{even}\ k \ge 0} ~ (-1)^{k/2} ~~ \sum_{\begin{smallmatrix} A \subseteq \{\,1,2,3,\dots\,\} \\ \left|A\right| = k\end{smallmatrix}} \left(\prod_{i \in A} \sin\theta_i \prod_{i \not \in A} \cos\theta_i\right) $ which isn't really enumerative, and I don't think there should be anything standing in the way of an enumerative expression in the case with finitely many terms. – Jackson May 16 '11 at 2:27
The formula you gave in the comment is slightly wrong, since you need to change both appearances of $\theta$ to $a$, and you need to allow $A$ to contain the number $0$. Also, please specify a definition of "closed-form expression" that doesn't make the problem tautological or impossible. For example, do you consider the expression $\sum_n a_n$ closed-form? – S. Carnahan May 16 '11 at 5:27
Yup, there are a couple typos above, but I don't see a way to edit that comment. I don't have a strict definition for "closed-form expression", and apparently that is a pretty vague term in general. This is as precise as I can make it: an expression with finitely many symbols denoting elementary operations such as addition, multiplication, etc. So no, I do not consider $\displaystyle\sum_{n=0}^\infty$ a closed-form expression as it involves a limiting process. As for the above expression involving permutation of subsets of natural numbers, I would reject it as not being explicit enough. – Jackson May 16 '11 at 8:46

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.

share|cite|improve this answer

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


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:


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:

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



Apologizes for the explosion of indices.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.