<s>[INST] 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
 [/INST] 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.
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