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UPSC NDA Math Model Paper 1

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Question : 94 of 120

Marks: +1, -0

If

\sin\alpha+\sin\beta+\sin\gamma=0

\cos\alpha+\cos\beta+\cos\gamma=0

, then the value of

\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)

is:

$-\frac{1}{2}$
$-\frac{3}{2}$
0
-1

Solution:

Given:

\sin\alpha+\sin\beta+\sin\gamma=0

and

\cos\alpha+\cos\beta+\cos\gamma=0

\sin\alpha+\sin\beta+\sin\gamma=0

\Rightarrow \sin\alpha+\sin\beta=-\sin\gamma

Sqauring both sides, we get

\Rightarrow \sin^{2}\alpha+\sin^{2}\beta+2\sin\alpha\sin\beta=\sin^{2}\gamma

.........(1)

And

\cos\alpha+\cos\beta+\cos\gamma=0

\Rightarrow \cos\alpha+\cos\beta=-\cos\gamma

Sqauring both sides, we get

\Rightarrow \cos^{2}\alpha+\cos^{2}\beta+2\cos\alpha\cos\beta

=\cos^{2}\gamma

......(2)

Adding (1) and (2) we get:

2+2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)=1

\Rightarrow 2+2\cos(\alpha-\beta)=1

\Rightarrow \cos(\alpha-\beta)=-\frac{1}{2}

Similarly,

\cos(\beta-\gamma)=-\frac{1}{2}

and

\cos(\gamma-\alpha)=-\frac{1}{2}

.

\therefore \cos(\alpha-\beta)+\cos(\beta-\gamma)

+\cos(\gamma-\alpha)=-\frac{3}{2} .

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