# An integral of composite function of triangle functions [closed]

I expected the following formula to hold:

$$\int^{2n\pi}_0\cos(\sin t+t/n)dt=0$$, for $${}^\forall n\in\mathbb{N},\ n\geq2$$

But I can't prove it. Could you please tell me.

• I'm not quite sure why people are voting to close this. Is it a well-known result? Is it obvious? I think the poster is saying that he found that this seems to be true numerically, and is asking for a proof or reference. I checked it numerically up to $n=10$. – Joe Silverman Jul 31 at 18:34
• I voted to close because I would have expected an explanation of why it might be expected to hold. More generally the question lacks context. Also it's evidently true for even $n$ by comparing $t$ and $t+n\pi$. – Anthony Quas Jul 31 at 18:45
• This is a question for MSE. – user64494 Jul 31 at 19:37

You can rewrite the integral as $$\int_0^{2\pi} \left(\sum_{j=0}^{n-1}\cos\Big(\sin t+\tfrac tn+2\pi \tfrac jn\Big)\right)\,dt.$$ But $$\sum_{j=0}^{n-1}\cos\big(a+2\pi \tfrac jn\big)=0$$ for all $$a$$.
In particular, the equality holds if $$\sin t$$ is replaced by any $$2\pi$$-periodic function.
• Nice. And for $n=1$ you sum is $\cos(a)$, not 0, which explains why the integral is non-zero for $n=1$. I guess there will be a difference of opinion as to whether this belongs on MO or MSE, but I will say it's a clever trick that I enjoyed seeing. – Joe Silverman Jul 31 at 20:30