What is the maximum value of $ \sin x \sin {2x}$ What is the maximum value of $$ \sin x \sin {2x}$$
I have done my work here 
$$f (x)=\sin x \sin 2x =\frac{\cos x - \cos3x}2 $$
$$f'(x)= \frac{- \sin x+3 \sin 3x}2 =4\sin x (2-3\sin^2 x)=0$$
$$x=0,\pi; \sin x= \sqrt{\frac {2}{3}}$$
$$f (0)=f (\pi)=0$$
$$f \left(\arcsin \sqrt{\frac{2}{3}}\right) =\frac{4}{3 \sqrt{3}}$$
If my work is not much right then please rectify it
 A: We have $$f(x) = \sin (x)\sin (2x) = 2{\sin ^2}(x)\cos (x) = 2(\cos (x) - {\cos ^3}(x))$$for a maximum we should have $f'(x) = 0$, so that $$ - \sin (x) + 3\sin (x){\cos ^2}(x) = 0$$so finally$$\left\{ \begin{array}{l}\sin (x) = 0 \to f(x) = 0\\\cos (x) = \frac{1}{{\sqrt 3 }} \to f(x) = \frac{4}{{3\sqrt 3 }}\\\cos (x) =  - \frac{1}{{\sqrt 3 }} \to f(x) =  - \frac{4}{{3\sqrt 3 }}\end{array} \right.$$which clearly shows the maximum.
A: Perhaps a more unorthodox approach: From the AM-GM inequality,
$$\sqrt[3]{\frac{1}{2}\sin^2x\cdot\frac{1}{2}\sin^2x\cdot\cos^2x} \le \frac{\frac{1}{2}\sin^2x\cdot + \frac{1}{2}\sin^2x\cdot + \cos^2x}{3}$$
$$\sqrt[3]{\frac{1}{4}\sin^4x\cos^2x} \le \frac{\sin^2x + \cos^2x}{3}$$
But it is well known that $\sin^2x + \cos^2x = 1$. Hence,
$$\sqrt[3]{\frac{1}{4}\sin^4x\cos^2x} \le \frac{1}{3}$$
$$\sin^4x\cos^2x\le\frac{4}{27}$$
Taking square roots of both sides,
$$-\frac{2}{\sqrt{27}} \le \sin^2x\cos x \le \frac{2}{\sqrt{27}}$$
Multiplying throughout by $2$,
$$-\frac{4}{\sqrt{27}} \le 2\sin^2x\cos x \le \frac{4}{\sqrt{27}}$$
But from the double angle formula, $2\sin^2x\cos x = \sin x \cdot2\sin x\cos x = \sin x \sin 2x$. Hence,
$$-\frac{4}{\sqrt{27}} \le \sin x \sin 2x \le \frac{4}{\sqrt{27}}$$
Equality for each bound occurs when $\frac{1}{2}\sin^2x = \cos^2x$, i.e. $\tan x = \pm \sqrt{2}$.
A: This is a slight simplification of Seyed's solution.
We have $$f(x) = \sin (x)\sin (2x) = 2(\cos (x) - {\cos ^3}(x))=2g(\cos(x))$$
for an extreme we should have 
$$0=g'(y) = 1-3 y^2,y=\cos(x)$$
so finally
$$ y = \frac{1}{\sqrt 3 } \to g(y) = \frac{4}{{3\sqrt 3 }}=f(x)$$
$$ y =  - \frac{1}{\sqrt 3 } \to g(y) =  - \frac{4}{{3\sqrt 3 }}=f(x)$$
which clearly shows the maximum.
A: It is false that $\sin x=0\implies x=0$ or $\pi$, and it is also false that $\sin x=\sqrt \frac 2 3 \implies x=\arcsin\sqrt \frac 2 3$. Do you see why, and can you explain why that's not a big deal?
