Question
Find an angle $\theta,0<\theta<\frac{\pi}{2},$ which increases twice as fast as its sine.

Answer

Let $\theta$ increases twice as fast as its sine
$\Rightarrow\ \theta=2\sin\theta$
$\Rightarrow\ \frac{\text{d}\theta}{\text{dt}}=2\cdot\cos\theta\cdot\frac{\text{d}\theta}{\text{dt}}$
$\Rightarrow\ 1=2\cos\theta$
$\Rightarrow\ \frac{1}{2}=\cos\theta$
$\Rightarrow\ \cos\theta=\cos\frac{\pi}{3}$
$\therefore\ \theta=\frac{\pi}{3}$
So, the required angle is $\frac{\pi}{3}$

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