# 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?

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 $-1\le t\le 1$.

• It is problems such as this one for which we have students spend time in pre-calculus courses solving "dreary" trigonometric equations... – colormegone Jun 12 '13 at 20:35

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 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

$$(**)\;\;\;=\cos^22\alpha\cos\alpha=\left(\cos^2\alpha-1\right)^2\cos\alpha$$

\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}