Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

This is a challenging puzzle I heard from my little brother.

For some $n$ and $x$, $\sum_{k=1}^n \sin^{2k}(x) = 2013$.

Is it possible to deduce $$\sum_{k=1}^n \cos^{2k}(x) \text{ ?}$$

Edit: I've just noticed something which now seems obvious to me.
Choose $n = 2013$ and $x = \pi/2$ which satisfies the condtion. It follows that the cosine terms would sum to zero. I'm not sure this is a unique solution.

share|cite|improve this question
Check for typos - the index $i$ doesn't seem to appear in the summand, which is usually a bad sign. Maybe it's supposed to be $\sum_{i=1}^n\sin^{2i}(x)$? – icurays1 Feb 12 '13 at 23:38
I've just noticed a trivial solution - x = $\pi/2$ and n = 2013. It follows that the cosine sum is 0. Maybe the problem is to show that this is the only solution? – Mark Feb 12 '13 at 23:41
For a large enough of $n$ there would be an $x$ satisfying the equation, by intermediate value theorem. – Maesumi Feb 12 '13 at 23:53
At $x=\pi/2$ the sum is $n$ and at $x=0$ the sum is $0$. The function, ie, the first sum, is a continuous function of $x$, so all values between 0 and $n$ will be be attained. Hence if $n\ge 2013$ then the value $2013$ will be attained. quite possibly multiple times. It looks unlikely that all such $x$'s will yield the same answer in the second sum. How about using $2$ instead of $2013$ and doing numerical experimentation. – Maesumi Feb 13 '13 at 0:00
@Mark Let $n \geq 2013$. Let $f(x)=\Sigma_{i=1}^{n} \sin^{2i}(x)$. Since $f(0)=0$ and $f(\frac{\pi}{2})>2013$ by the IVT there exists some $x$ so that $f(x)=2013$. – N. S. Feb 13 '13 at 0:01
up vote 5 down vote accepted

let $r=sin^2(x)$ we have

$$\sum_{k=1}^n r^k=\frac{r(1-r^n)}{1-r}=2013$$ Now we want: $$\sum_{k=1}^n (1-r)^k=\frac{(1-r)(1-(1-r)^n)}{r}$$

We can deduce: $$2013\sum_{k=1}^n (1-r)^k=(1-r^n)(1-(1-r)^n)$$

share|cite|improve this answer
So, if you know $n$, then you can evaluate the cosine sum. – Gerry Myerson Feb 12 '13 at 23:58
@GerryMyerson: I have not tried yet. but the second sum may be simplified and obtained from first one. – user59671 Feb 13 '13 at 0:01
If you know $n$ (and if you can solve a polynomial equation of high degree), yes. If you don't know $n$, I think my answer shows that you can't find the cosine sum just from knowing the sine sum. – Gerry Myerson Feb 13 '13 at 0:25
@GerryMyerson: It seems the second sum cannot be written in terms of first sum free from n. And when $n$ is given, one may not need to find an exact solution for $r$ and then put it in second one. for example if one can show the first equation is symmetric around 1/2 he may be able to find the second. However it doesn't seem to be symmetric around 1/2... – user59671 Feb 13 '13 at 0:33
@Mark: Suppose r and s are two solutions and s<r. Then $$\sum_{k=1}^n{r^k-s^k}=0$$ which is impossible. – user59671 Feb 13 '13 at 19:11

As noted, the equation holds if $n=2013$ and $x=\pi/2$. Now let $n=2014$. By continuity, there is a value of $x$ a tiny bit smaller than $\pi/2$ for which the equation will hold, and, for this value of $x$, the cosine sum will not be zero. So one cannot deduce the cosine sum from knowing the first equation holds.

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.