Question

Prove that $\theta =\pi /3$ at $\sin \theta (1+\cos \theta )$ are maximum.

Answer 1

$\begin{array}{c}f\left(\theta \right)=\sin \theta \left(1+\cos \theta \right)\\ f\left(\theta \right)=\sin \theta +\frac{1}{2}\sin 2\theta \\ f^{\prime }\left(\theta \right)=\cos \theta +\frac{1}{2}\times 2\sin 2\theta \\ f^{\prime }\left(\theta \right)=\cos \theta +\cos 2\theta \\ ={\cos }^2\theta -\left(1-{\cos }^2\theta \right)+\cos \theta \\ =2{\cos }^2\theta +\cos \theta +1\\ =2\cos \theta \left(\cos \theta +1\right)-\left(\cos \theta +1\right)\\ =\left(2\cos \theta -1\right)\left(\cos \theta +1\right)\\ f^{\prime }\left(\theta \right)=0\\ \cos \theta =\frac{1}{2}\cos \theta =-1\\ \theta =\frac{\pi }{3},\pi ,\frac{5\pi }{3}\\ \underset{–}{(+)(-)(+)}\\ \frac{\pi }{3}\pi \frac{5\pi }{3}\end{array}$

At $x=\frac{\pi }{3}{f}^{\prime }\left(\theta \right)$ changes sign from $(+ve)$ to $(-ve)$.

$\therefore$ At $\theta =\frac{\pi }{3}$ $f\left(\theta \right)$ is maximum.