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Questions  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Q)

If $\cos\alpha+\cos\beta=0=\sin\alpha+\sin \beta$,then prove that $\cos 2\alpha+\cos 2\beta=-2\cos(\alpha+\beta)$

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  • $\cos(a+b)=\cos a\cos b-\sin a \sin b$
  • $\cos 2\theta=\cos^2\theta-\sin^2\theta$
Given :
$\cos\alpha+\cos\beta=0$
Squaring both side
$(\cos\alpha+\cos\beta)^2=0$
$\Rightarrow \cos^2\alpha+\cos^2\beta+2\cos \alpha.\cos\beta=0$
$\Rightarrow \cos^2\alpha+\cos^2\beta=-2\cos \alpha.\cos \beta$------(1)
Similarly
$\sin\alpha+\sin\beta=0$
$\Rightarrow \sin^2\alpha+\sin^2\beta=-2\sin \alpha.\sin \beta$------(2)
L.H.S
$\cos 2\alpha+\cos 2\beta$
$\cos ^2\alpha-\sin^ 2\alpha+\cos 2\beta-\sin^2\beta$
$(\cos ^2\alpha+\cos 2\beta)-(\sin^2\alpha+\sin^2\beta)$
From (1) & (2)
$-2\cos \alpha\cos\beta+2\sin \alpha\sin \beta$
$-2(\cos \alpha\cos\beta-\sin \alpha\sin \beta)$
$-2\cos(\alpha+\beta)$=R.H.S
Hence proved.
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