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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Prove :$\cos 6x=32\cos ^6x-48\cos^4x+18\cos^2x-1$

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Toolbox:
  • $\cos 3x=4\cos^3x-3\cos x$
  • $\cos 2x=2\cos^2x-1$
L.H.S
$\cos 6x=\cos 3(2x)$
$\Rightarrow 4\cos^32x-3\cos 2x$
$\Rightarrow 4(2\cos^2x-1)^3-3(2\cos^2x-1)$
$\Rightarrow 4(8\cos^6x-12\cos^4x+6\cos^2x-1)-6\cos^2x+3$
$\Rightarrow 32\cos^6x-48\cos^4x+24\cos^2x-4-6\cos^2x+3$
$\Rightarrow 32\cos ^6x-48\cos^4x+18\cos^2x-1$=R.H.S
Hence proved.
answered Apr 17, 2014 by sreemathi.v
 
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