Maximum value of a function. I am not able to check double derivative. Can someone explain me how $\sin^p x \cos^q x$ attains maximum at $\tan^2 x = \frac pq$.
I am not able to check whether double derivative is positive or negative.

Question
Show that $$\sin^p\theta\cos^q\theta$$ attains a maximum when $$\theta=\tan^{-1}\sqrt{(p/q)}$$
Solution
Ley $y=\sin^p\theta\cos^q\theta$. For a maximum or minimum of $y$, we have $\frac{\text dy}{\text dx}=0$
\begin{align}p\sin^{p-1}\theta\cos^{q+1}\theta-q\sin^{p+1}\theta\cos^{q-1}\theta&=0\\
\sin^{p-1}\theta\cos^{q-1}\theta(p\cos^2\theta-q\sin^2\theta)&=0\end{align}
Therefore \begin{align}\sin\theta&=0\\
&\Downarrow\\
\theta&=0\\
\text{or } \cos\theta&=0\\
&\Downarrow\\
\theta&=\frac \pi 2\\
\text{or }\tan^2\theta&=\frac pq\\
&\Downarrow\\
\theta&=\tan^{-1}\sqrt{p/q}\end{align}
Now $y=0$ at $\theta=0$ and also at $\theta=\frac\pi2$
When $0<\theta<\frac\pi2$, $y$ is positive
Also, $\tan^{-1}\sqrt{p/q}$ is the only value of $\theta$ lying between $0$ and $\frac \pi2$ at which $\frac{\text d}{\text dx}=0$.
Hence $y$ is maximum when $$\theta=\tan^{-1}\sqrt{p/q}$$
This can be seen from the graph of $y$


Let $f(x) = \sin^{p}(x) \, \cos^{q}(x)$ then
\begin{align}
f'(x) &= f(x) \, (-q \, \tan(x) + p \, \cot(x)) \\
&= -q \, \cot(x) \, f(x) \, \left( \tan^{2}(x) - \frac{p}{q} \right)
\end{align}
Now, the maximum of a function is defined by $f'(x) = 0$ then, seeking values other than $f(x) = 0$,
$$\tan^{2}(x) = \frac{p}{q}$$
or 
$$x_{max} = \tan^{-1}\left(\sqrt{\frac{p}{q}}\right).$$
