I want to show that the following inequality is true when $x$ is on the interval $[0,\frac{\pi}{2}]$:

$$\sin(nx)\cos(x)+\sin(x)\cos(nx) \le \sin(nx)+\sin(x)$$

I believe it to be true just from crunching numbers, but I would like an elegant way to show it. I am really new to doing proofs, so I am not sure how to proceed.

Thank you!

  • $\begingroup$ Since it's $\cos(nx)$, the LHS becomes $\sin(nx+x)$. $\endgroup$
    – user236182
    Oct 6 '15 at 16:53
  • $\begingroup$ @Dr.MV - there is still an issue, though. $\sin(nx)$ and $\cos(nx)$ could be negative, depending on $n$. $\endgroup$ Oct 6 '15 at 16:56
  • $\begingroup$ But the interval was between 0 and pi/2, and n is in the natural numbers, so no they can't be negative. $\endgroup$
    – Physika
    Oct 6 '15 at 16:57
  • $\begingroup$ @PaulSinclair Yes, we would need absolute values to make this trivial. $\endgroup$
    – Mark Viola
    Oct 6 '15 at 17:00
  • $\begingroup$ My end goal is to make the LHS look like (n+1)sin(x). $\endgroup$
    – Physika
    Oct 6 '15 at 17:02

This is not a valid inequality. Consider $n=4$ then plot $\sin 5 x - \sin 4x-\sin x$, we obtain

enter image description here

  • $\begingroup$ Interesting, when plotting the difference for high values of $n$, you see a kind of envelope, which is slowly increasing. It seems to stay always below $0.5$, so that would mean that adding $1/2$ on the right would render the inequality true. $\endgroup$ Oct 6 '15 at 17:23
  • $\begingroup$ @Jean-ClaudeArbaut this is true ... but I dropped the idea of correcting the inequality :-) $\endgroup$
    – Math-fun
    Oct 6 '15 at 18:47
  • $\begingroup$ Ah ok, thank you very much for your help. So it seems I couldn't prove it because it is not true! $\endgroup$
    – Physika
    Oct 7 '15 at 17:10

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