Sine Series

Geometry Level 3

$\LARGE y=sin\theta\sqrt{sin\theta\sqrt{sin\theta......}}$

If $y\neq0$ . For whice value of $\theta$ , $y=\frac{1}{2}$

\frac{\pi}{3}

\pi

\frac{\pi}{8}

0

\frac{\pi}{6}

None of them

\frac{\pi}{4}

\frac{\pi}{2}

1 solution

Md Mehedi Hasan
Oct 29, 2017

$y=sin\theta\sqrt{sin\theta\sqrt{sin\theta......}}\\\Rightarrow y=sin\theta\sqrt{y}\\\Rightarrow y^2=sin^2\theta\times{y}\\\Rightarrow y(y-sin^2\theta)=0\\\Rightarrow y=sin^2\theta { \quad \quad \quad \quad {\color{#3D99F6}\boxed{y\neq0}}}\\\Rightarrow \frac{1}{2}=sin^2\theta\\\Rightarrow sin\theta=\frac{1}{\sqrt2}\\\therefore \theta=\boxed{\frac{\pi}{4}}$

Sine Series

1 solution

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