Question

Find the value of $\theta$ for attaining a max value of ${\sin }^p\theta {\cos }^q\theta$ is

• A. ${\tan }^{-1}\frac{p}{q}$
• B. ${\tan }^{-1}\frac{\sqrt{p}}{\sqrt{q}}$
• C. ${\tan }^{-1}(p+q)$
• D. ${\tan }^{-1}q$

Answer 1

The correct option is B ${\tan }^{-1}\frac{\sqrt{p}}{\sqrt{q}}$

The given function is:

$y={\sin }^p\theta {\cos }^q\theta$

Differentiating once we get.

$\Rightarrow y^{\prime }=p{\sin }^{p-1}\theta {\cos }^{q+1}\theta -q{\sin }^{p+1}\theta {\cos }^{q-1}\theta$

Equating $y^{\prime }=0$ to find the extremum value we get,

$\Rightarrow p{\sin }^{p-1}\theta {\cos }^{q+1}\theta -q{\sin }^{p+1}\theta {\cos }^{q-1}\theta =0$

$\Rightarrow \frac{p}{q}=\frac{{\sin }^2\theta }{{\cos }^2\theta }$

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

$\Rightarrow \tan \theta =\sqrt{\frac{p}{q}}$

$\tan \theta =\frac{\sqrt{p}}{\sqrt{q}}$

$\theta ={\tan }^{-1}\frac{\sqrt{p}}{\sqrt{q}}$ .....Answer