Question:
Find an angle q, 0 < q < /2, which increases twice as fast as its sine.
Solution:
According to the question, we have
$\frac{d \theta}{d t}=2 \frac{d}{d t}(\sin \theta)$
$\Rightarrow \frac{d \theta}{d t}=2 \cos \theta \cdot \frac{d \theta}{d t} \Rightarrow 1=2 \cos \theta$
So, $\cos \theta=\frac{1}{2} \Rightarrow \cos \theta=\cos \frac{\pi}{3} \Rightarrow \theta=\frac{\pi}{3}$
Therefore, the required angle is π/3.