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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Prove : $\cos 4x=1-8\sin^2x\cos^2x$

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  • $\cos 2x=2\cos^2x-1$
L.H.S
$\cos 2(2x)=2\cos^22x-1$
$\Rightarrow 2(2\cos x-1)^2-1$
$\Rightarrow 2(4\cos^2x+1-4\cos x)-1$
$\Rightarrow 8\cos^4x+2-8\cos^2x-1$
$\Rightarrow 8\cos^4x-8\cos^2x+1$
$\Rightarrow 8\cos^2x(\cos^2x-1)+1$
$\Rightarrow 8\cos^2x(\sin^2x)+1$
$\Rightarrow 1+8\cos^2x.\sin^2x$=R.H.S
Hence proved.
answered Apr 17, 2014 by sreemathi.v
 
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