Find the maximum value of $r$ when $r=\cos\alpha \sin2\alpha$ 
Find the maximum value of $r$ when $$r=\cos\alpha \sin2\alpha$$ 

$$\frac{\rm dr}{\rm d\alpha}=(2\cos2\alpha )(\cos\alpha)-(\sin2\alpha)(\sin\alpha)=0 \tag {at maximum}$$
How do I now find alpha?
 A: We have 
$$\cos\alpha\sin 2\alpha=\cos\alpha(2\sin\alpha\cos\alpha)=2(1-\sin^2\alpha)\sin\alpha=2\sin\alpha-2\sin^3\alpha.$$
Examine the function $t-t^3$, where $-1\le t\le 1$. 
A: You are better off just expressing everything in terms of $\sin{\alpha}$:
$$\cos{\alpha} \sin{2 \alpha} = 2 \sin{\alpha} \cos^2{\alpha} = 2 \sin{\alpha} - 2 \sin^3{\alpha}$$
Can you take it from here?
A: A trigonometric idea:
$$(**)\;\;\;\;2\cos2\alpha\cos\alpha-\sin2\alpha\sin\alpha=\cos\alpha\cos2\alpha\cos3\alpha$$
since
$$\cos(x+y)=\cos x\cos y-\sin x\sin y$$
so in fact, using the formula for product of cosines
$$(**)\;\;\;=\cos^22\alpha\cos\alpha=\left(\cos^2\alpha-1\right)^2\cos\alpha$$
A: $$
\begin{align}
r&=\cos(\alpha)\sin(2\alpha)\\
\frac{\mathrm{d}r}{\mathrm{d}\alpha}&=2\cos(\alpha)\cos(2\alpha)-\sin(\alpha)\sin(2\alpha)
\end{align}
$$
Setting $\frac{\mathrm{d}r}{\mathrm{d}\alpha}=0$ yields
$$
\begin{align}
2&=\tan(\alpha)\tan(2\alpha)\\
&=\frac{2\tan^2{\alpha}}{1-\tan^2{\alpha}}\\
\tan^2(\alpha)&=1/2\\
\sin^2(\alpha)&=1/3\\
\cos^2(\alpha)&=2/3
\end{align}
$$
Plugging this back into $r$ yields
$$
\begin{align}
r
&=\cos(\alpha)\sin(2\alpha)\\
&=2\sin(\alpha)\cos^2(\alpha)\\
&=\frac4{3\sqrt3}
\end{align}
$$
