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UPSC NDA I 2016 Mathematics Solved Paper

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For the next two (2) items that follow :

Consider a triangle ABC in which

\cos A+\cos B+\cos C=\sqrt{3}\sin\frac{\pi}{3}

[NDA I 2016]

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Question : 36 of 120

Marks: +1, -0

What is the value of

\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \;\;?

$\frac{1}{2}$
$\frac{1}{4}$
$\frac{1}{8}$
$\frac{1}{16}$

Solution:

Given

\cos A+\cos B+\cos C=\sqrt{3}\sin\;\;\frac{\pi}{3}

\cos A+\cos B+\cos C

=2\cos\;\;\frac{A+B}{2}\cdot\cos\;\;\frac{A-B}{2}+\left(1-2\sin^{2}\;\;\frac{C}{2}\right)

=2\sin\;\;\frac{C}{2}\cos\;\;\frac{A-B}{2}+1-2\sin^{2}\;\;\frac{C}{2}

\left(\because\;\;\frac{A+B}{2}=90^{\circ}-\;\;\frac{C}{2}\right)

=1+2\sin\;\;\frac{C}{2}\left[\cos\;\;\frac{A-B}{2}-\sin\;\;\frac{C}{2}\right]

=1+2\sin\;\;\frac{C}{2}\left[\cos\;\;\frac{A-B}{2}-\cos\;\;\frac{A+B}{2}\right]

=1+4\sin\;\;\frac{C}{2}\sin\;\;\frac{A}{2}\sin\;\;\frac{B}{2}

\cos A+\cos B+\cos C=\sqrt{3}\sin\;\;\frac{\pi}{3}

1+4\sin\;\;\frac{C}{2}\sin\;\;\frac{A}{2}\sin\;\;\frac{B}{2}=\sqrt{3}\sin\;\;\frac{\pi}{3}

\Rightarrow 1+4\sin\;\;\frac{C}{2}\sin\;\;\frac{A}{2}\sin\;\;\frac{B}{2}=\sqrt{3}\times\;\;\frac{\sqrt{3}}{2}

\Rightarrow 4\sin\;\;\frac{A}{2}\sin\;\;\frac{B}{2}\sin\;\;\frac{C}{2}=\;\;\frac{3}{2}-1

\sin\;\;\frac{A}{2}\sin\;\;\frac{B}{2}\sin\;\;\frac{C}{2}=\;\;\frac{1}{8}

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