# Showing $\left\lvert \frac{\sin(nx)}{n\sin(x)} + \frac{\cos(nx)}{n\cos(x)} \right\rvert \le\left\lvert\frac{n+1}{n}\right\rvert$

It was shown in here that $$\left\lvert \frac{\sin(nx)}{n\sin(x)}\right\rvert \le1\,\,\forall x\in\mathbb{R}-\{\pi k: k\in\mathbb{Z}\}$$ iff $n$ is a non-zero integer.

Using the similar argument in the same post, we are able to show that $$\left\lvert \frac{\cos(nx)}{n\cos(x)}\right\rvert \le1\,\,\forall x\in\mathbb{R}-\{\frac{(2k+1)\pi}{2} : k\in\mathbb{Z}\}$$ iff $n$ is an odd non-zero integer (Please alert me if this is wrong).

Now, by some graph sketching, it seems that $$\left\lvert \frac{\sin(nx)}{n\sin(x)} + \frac{\cos(nx)}{n\cos(x)} \right\rvert \le\left\lvert \frac{n+1}{n}\right\rvert \,\,\forall x\in\mathbb{R}-\{\frac{k\pi}{2} : k\in\mathbb{Z}\}$$ iff $n$ is an odd non-zero integer.

I am not sure if the above inquality is true. Please enlighten me!

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If it is true, we may write $$\left\lvert \frac{\sin(nx)}{\sin(x)} + \frac{\cos(nx)}{\cos(x)} \right\rvert \le\left\lvert n+1\right\rvert \,\,\forall x\in\mathbb{R}-\{\frac{k\pi}{2} : k\in\mathbb{Z}\}$$ iff $n$ is an odd non-zero integer, instead. –  pipi Oct 22 '12 at 5:49

You could make progress as follows:$$\frac{\sin(nx)}{\sin(x)} + \frac{\cos(nx)}{\cos(x)}=2\frac{\sin(nx)\cos(x)+\cos(nx)\sin(x)}{2\cos(x)\sin(x)}=\frac{2\sin((n+1)x)}{\sin(2x)}$$

Then with $y=2x$ you find that this is equal to $\cfrac {2\sin(\frac{(n+1)y}{2})}{\sin(y)}$ and you just need to jiggle with the factors to use the original result [odd $n$ gives an integer]

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@ Mark Bennet I manage to get it after I posted the question! Anyway, it is a nice result to share! It may be an extension of the previous question... –  pipi Oct 22 '12 at 6:21

Multiplying everything by $n\sin(x)\cos(x)$ and using the addition formula for sines, one sees that the task is to prove $|\sin((n+1)x)|\leqslant|n+1|\cdot|\sin(x)\cos(x)|$. Since $n$ is odd, $n=2k-1$ and, replacing $x$ by $\frac12x$, one is left with $|\sin(kx)|\leqslant |k|\cdot|\sin(x)|$.

The last inequality depends only on $|k|$ and it obviously holds for $k=0$ and $k=1$ hence a recursion over $k\geqslant1$ shows that it holds for every integer $k$.

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$$\frac{2\sin((n+1)x)}{n\sin(2x)}\frac{2\sin((n+1)x)}{n\sin(2x)}\frac{\sin(nx)\cos(x) + \cos(nx)\sin(x)}{n\sin(x)\cos(x)} =\frac{2\sin((n+1)x)}{n\sin(2x)}=\frac{2(n+1))}{n} \frac{\sin((n+1)x)}{\frac{n+1}{2}\sin(2x)}$$

If $n+1$ is even, this inequality follows immediately from the mentioned one, while if $n+1$ is odd you should be able to prove it exactly the same way as the stated one. Actually, since you need to prove it anyhow, show directly that

$$\left| \frac{\sin((n+1)x)}{\frac{n+1}{2}\sin(2x)} \right| \leq 1$$

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