Exercise 8.8

Since the functions on both sides of the inequality are periodic functions with period $\pi$, and since they are both even functions, we only have to show this for $x\in [0,\pi ∕2]$.

Let $f(x)=n\sin <!--\; FUNCTION\; APPLICATION\; -->x-\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}$, $f(0)=0$, $f^{\prime }(x)=n(\cos <!--\; FUNCTION\; APPLICATION\; -->x-\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx})$. Since $\cos <!--\; FUNCTION\; APPLICATION\; -->x$ is monotonically decreasing on $[0,\pi ∕2]$, we have $f^{\prime }(x)\ge 0$ for $x\in [0,\pi ∕(2n)]$, hence $f(x)\ge 0$ in this interval, that is, $n\sin <!--\; FUNCTION\; APPLICATION\; -->x\ge \sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}$.

For $x\in [\pi ∕(2n),\pi ∕2]$, since $\sin <!--\; FUNCTION\; APPLICATION\; -->x$ is monotonically increasing, we have $n\sin <!--\; FUNCTION\; APPLICATION\; -->x\ge n\sin <!--\; FUNCTION\; APPLICATION\; -->(\pi ∕(2n))\ge \sin <!--\; FUNCTION\; APPLICATION\; -->(\mathit{n\pi }∕(2n))=1$, while $|\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}|\le 1$. Hence $n\sin <!--\; FUNCTION\; APPLICATION\; -->x\ge |\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}|$ on all of $[0,\pi ∕2]$.