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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.

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

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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.

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