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UPSC NDA Math Model Paper 5

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Question : 14 of 120

Marks: +1, -0

The value of

\sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}

is

$\frac{1}{2}$
$\frac{\sqrt{3}}{2}$
$\frac{\sqrt{3}}{8}$
$\frac{1}{8}$

Solution:

S = sin 20° sin 40° sin 80°

\Rightarrow S=\frac{1}{2}\sin 40^{\circ}(2\sin 80^{\circ}\sin 20^{\circ})

\Rightarrow S=\frac{1}{2}\sin 40^{\circ}[\cos(80-20)-\cos(80+20)]

\Rightarrow S=\frac{1}{2}\sin 40^{\circ}[\cos 60-\cos 100]

\Rightarrow S=\frac{1}{4}\sin 40^{\circ}-\frac{1}{2}\sin 40\cos 100

\Rightarrow S=\frac{1}{4}\sin 40^{\circ}-\frac{1}{4}(2\sin 40\cos 100)

\Rightarrow S=\frac{1}{4}\sin 40^{\circ}-\frac{1}{4}[\sin(100+40)

-\sin(100-40)]

\Rightarrow S=\frac{1}{4}\sin 40^{\circ}-\frac{1}{4}[\sin 140-\sin 60]

\Rightarrow S=\frac{1}{4}\sin 40^{\circ}-\frac{1}{4}\sin 140+\frac{1}{4}\sin 60

\Rightarrow \frac{1}{4}(\sin 40^{\circ}-\sin 140)+\frac{\sqrt{3}}{8}

\Rightarrow S=\frac{1}{4}[\sin 40^{\circ}-\sin(180-40)]+\frac{\sqrt{3}}{8}

\because \sin(180-x)=\sin x

\Rightarrow S=\frac{1}{4}(\sin 40^{\circ}-\sin 40)+\frac{\sqrt{3}}{8}

\Rightarrow S=\frac{\sqrt{3}}{8}

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