Question

The angle $\mathrm{\theta },0<\mathrm{\theta }<\frac{\mathrm{\pi }}{2},$ which increases twice as fast as its sine, is _________________.

Answer 1

Let angle θ increase twice as fast as its sine.

It is given that,

$\frac{d\theta }{dt}=2\frac{d}{dt}\left(\sin \theta \right)$

$\Rightarrow \frac{d\theta }{dt}=2\cos \theta \frac{d\theta }{dt}$

$\Rightarrow 2\cos \theta =1$

$\Rightarrow \cos \theta =\frac{1}{2}=\cos \frac{\mathrm{\pi }}{3}$

$\Rightarrow \theta =\frac{\mathrm{\pi }}{3}\left(0<\theta <\frac{\mathrm{\pi }}{2}\right)$

Thus, the angle θ is $\frac{\mathrm{\pi }}{3}$.


The angle $\mathrm{\theta },0<\mathrm{\theta }<\frac{\mathrm{\pi }}{2},$ which increases twice as fast as its sine, is $\underline{\frac{\mathrm{\pi }}{3}}$.