Question:
The angle $\theta, 0<\theta<\frac{\pi}{2}$, which increases twice as fast as its sine, is ______________
Solution:
Let angle θ increase twice as fast as its sine.
It is given that,
$\frac{d \theta}{d t}=2 \frac{d}{d t}(\sin \theta)$
$\Rightarrow \frac{d \theta}{d t}=2 \cos \theta \frac{d \theta}{d t}$
$\Rightarrow 2 \cos \theta=1$
$\Rightarrow \cos \theta=\frac{1}{2}=\cos \frac{\pi}{3}$
$\Rightarrow \theta=\frac{\pi}{3}$ $\left(0<\theta<\frac{\pi}{2}\right)$
Thus, the angle $\theta$ is $\frac{\pi}{3}$.
The angle $\theta, 0<\theta<\frac{\pi}{2}$, which increases twice as fast as its sine, is $\frac{\pi}{3}$
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