# Find an angle $\theta$,$0$ <$\;\theta<\Large \frac{1}{2}$,which increases twice as fast as its sine.

$\begin{array}{1 1} \frac{\pi}{6} \\ \frac{\pi}{3} \\ \frac{\pi}{4} \\ \frac{\pi}{2} \end{array}$

Toolbox:
• If $y = f(x)$, then $\large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $x$.
• $\bigg(\large \frac{dy}{dx} \bigg)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$
Step 1
Let $x = \sin\: \theta$
On differentiating w.r.t $t$ we get,
$\large\frac{dx}{dt}$$=\cos\theta\large\frac{d\theta}{dt} But it is given that \large\frac{d\theta}{dt}$$=2.\large\frac{dx}{dt}$
$\Rightarrow \large\frac{dx}{dt}$$=\cos\theta.2\large\frac{dx}{dt}$
$\Rightarrow \cos\theta=\large\frac{1}{2}$
$\Rightarrow \theta = \large\frac{\pi}{3}$
Hence the angle is $\large\frac{\pi}{3}$
edited Feb 5, 2014