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

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Given: a condition \theta, 0<\theta<\frac{\pi}{2}

To find:  the angle θ such that it increases twice as fast as its sine.

Explanation: Let x = sin θ

Let’s differentiate with respect to t,

\frac{d x}{d t}=\frac{d(\sin \theta)}{d t}
Now applying the derivative results,

\frac{d x}{d t}=\cos \theta \cdot \frac{d(\theta)}{d t} \ldots . .(i)
As this is given in the question

\frac{\mathrm{d}(\theta)}{\mathrm{dt}}=2 \cdot \frac{\mathrm{d} x}{\mathrm{dt}}

Let's substitute this value in equation (i)

\frac{d x}{d t}=\cos \theta \cdot 2 \cdot \frac{d x}{d t}

After cancelling the like terms, we get 1=2 \cos \theta
\Rightarrow \cos \theta=\frac{1}{2}
But given 0<\theta<\frac{\pi}{2}, this possibility occurs only when  \cos \theta=\frac{\pi}{3}
Hence the angle \theta is \frac{\pi}{3}.

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