# Show that $\sin(nx)\cos(x)+\sin(x)\cos(nx)\le \sin(nx)+\sin(x)$??

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!

• Since it's $\cos(nx)$, the LHS becomes $\sin(nx+x)$. Oct 6 '15 at 16:53
• @Dr.MV - there is still an issue, though. $\sin(nx)$ and $\cos(nx)$ could be negative, depending on $n$. Oct 6 '15 at 16:56
• But the interval was between 0 and pi/2, and n is in the natural numbers, so no they can't be negative. Oct 6 '15 at 16:57
• @PaulSinclair Yes, we would need absolute values to make this trivial. Oct 6 '15 at 17:00
• My end goal is to make the LHS look like (n+1)sin(x). 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 • 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. Oct 6 '15 at 17:23