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Prove : $\sin(n+1)x\sin(n+2)x+\cos(n+1)x\cos(n+2)x=\cos x$

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  • $\cos(A-B)=\cos A.\cos B+\sin A\sin B$
L.H.S
$\sin(n+1)x\sin(n+2)x+\cos(n+1)x\cos(n+2)x$
Put $(n+2)x=A,(n+1)x=B$
$\sin A.\sin B+\cos A\cos B=\cos(A-B)$
$\therefore \cos(A-B)=\cos[(n+2)x-(n+1)x]$
$\Rightarrow \cos(nx+2x-nx-x)$
$\Rightarrow \cos x$=R.H.S
Hence proved.
answered Apr 17, 2014 by sreemathi.v
 

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