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# If $\sin(\theta+\alpha)=a$ and $\sin(\theta+\beta)=b$,then prove that $\cos 2(\alpha-\beta)-4ab\cos(\alpha-\beta)=1-2a^2-2b^2$

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A)
Toolbox:
• $\cos(a-b)=\cos a\cos b+\sin a\sin b$
• $\cos 2\theta=2\cos^2\theta-1$
• $\cos^2\theta=1-\sin^2\theta$
Given :
$\sin(\theta+\alpha)=a$
$\sin(\theta+\beta)=b$
$\cos2(\alpha-\beta)-4ab\cos(\alpha-\beta)=1-2a^2-2b^2$
Express $\cos(\alpha-\beta)=\cos[(\theta+\alpha)-(\theta+\beta)]$
$\Rightarrow \cos(\theta+\alpha).\cos(\theta+\beta)+\sin(\theta+\alpha).\sin(\theta+\beta)$
$\Rightarrow \cos(\theta+\alpha).\cos(\theta+\beta)+a.b$-----(1)
L.H.S
$\cos2(\alpha-\beta)-4ab\cos(\alpha-\beta)$
$2\cos^2(\alpha-\beta)-1-4ab\cos(\alpha-\beta)$
From (1) we have
$2[\cos(\theta+\alpha).\cos(\theta+\beta+ab]^2-1-4ab[\cos(\theta+\alpha).\cos(\theta+\beta)+ab]$
$2[\cos^2(\theta+\alpha).\cos^2(\theta+\beta)+a^2b^2+2ab\cos(\theta+\beta).\cos(\theta+\alpha)]-1-4ab\cos(\theta+\alpha).\cos(\theta+\beta)-4a^2b^2$
$2\cos^2(\theta+\alpha).\cos^2(\theta+\beta)+2a^2b^2+4ab\cos(\theta+\beta).\cos(\theta+\alpha)-1-4ab\cos(\theta+\alpha).\cos(\theta+\beta)-4a^2b^2$
$2[(1-\sin^2(\theta+\alpha)(1-\sin^2(\theta+\beta)]-1-2a^2b^2$
$2[(1-a^2)(1-b^2)-1-2a^2b^2$
$2[1-b^2-a^2+a^2b^2]-1-2a^2b^2$
$2-2b^2-2a^2+2a^2b^2-1-2a^2b^2$
$1-2a^2-2b^2$=R.H.S
Hence proved.