Class 11MathematicsTrigonometric FunctionsTrigonometrical ratios of multiple and sub-multiple angles
MediumProve that: $\cos 6x = 32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x - 1$
$L.H.S. = \cos 6x$
$= \cos 3(2x)$
$= 4 \cos^3 2x - 3 \cos 2x$ (using $\cos 3A = 4 \cos^3 A - 3 \cos A$)
$= 4[(2 \cos^2 x - 1)^3 - 3(2 \cos^2 x - 1)]$ (using $\cos 2x = 2 \cos^2 x - 1$)
$= 4[8 \cos^6 x - 1 - 3(4 \cos^4 x)(1) + 3(2 \cos^2 x)(1)^2] - 6 \cos^2 x + 3$
$= 4[8 \cos^6 x - 12 \cos^4 x + 6 \cos^2 x - 1] - 6 \cos^2 x + 3$
$= 32 \cos^6 x - 48 \cos^4 x + 24 \cos^2 x - 4 - 6 \cos^2 x + 3$
$= 32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x - 1$
$= R.H.S.$
$= \cos 3(2x)$
$= 4 \cos^3 2x - 3 \cos 2x$ (using $\cos 3A = 4 \cos^3 A - 3 \cos A$)
$= 4[(2 \cos^2 x - 1)^3 - 3(2 \cos^2 x - 1)]$ (using $\cos 2x = 2 \cos^2 x - 1$)
$= 4[8 \cos^6 x - 1 - 3(4 \cos^4 x)(1) + 3(2 \cos^2 x)(1)^2] - 6 \cos^2 x + 3$
$= 4[8 \cos^6 x - 12 \cos^4 x + 6 \cos^2 x - 1] - 6 \cos^2 x + 3$
$= 32 \cos^6 x - 48 \cos^4 x + 24 \cos^2 x - 4 - 6 \cos^2 x + 3$
$= 32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x - 1$
$= R.H.S.$
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