Question

Prove that the extremum value of the function ${\sin }^p\theta {\cos }^q\theta$ is at $\tan \theta =\pm \left(\sqrt{\frac{p}{q}}\right)$.

Answer 1

Let $f\left(\theta \right)=$${\sin }^p\theta {\cos }^q\theta$.

Now, $f^{\prime }\left(\theta \right)=p{\sin }^{p-1}\theta .{\cos }^{q+1}\theta -q{\sin }^{p+1}\theta .{\cos }^{q-1}\theta$.

For extremum value of $f\left(\theta \right)$ we must have $f^{\prime }\left(\theta \right)=0$

or, ${\sin }^{p-1}\theta .{\cos }^{q-1}\theta (p{\cos }^2\theta -q{\sin }^2\theta )=0$

or, ${\tan }^2\theta =\frac{p}{q}$

or, $\tan \theta =\pm \sqrt{\frac{p}{q}}$.

So it is clear the function $f\left(\theta \right)$ has extremum at $\tan \theta =\pm \sqrt{\frac{p}{q}}$.