Class 11MathematicsTrigonometrical Ratios, Functions and IdentitiesTrigonometrical ratios of multiple and sub-multiple angles
MediumProve that: $\cos 4x = 1 - 8 \sin^2 x \cos^2 x$
(N/A) $L.H.S. = \cos 4x$
$= \cos 2(2x)$
$= 1 - 2 \sin^2 2x$ (using $\cos 2A = 1 - 2 \sin^2 A$)
$= 1 - 2(2 \sin x \cos x)^2$ (using $\sin 2A = 2 \sin A \cos A$)
$= 1 - 2(4 \sin^2 x \cos^2 x)$
$= 1 - 8 \sin^2 x \cos^2 x$
$= R.H.S.$
$= \cos 2(2x)$
$= 1 - 2 \sin^2 2x$ (using $\cos 2A = 1 - 2 \sin^2 A$)
$= 1 - 2(2 \sin x \cos x)^2$ (using $\sin 2A = 2 \sin A \cos A$)
$= 1 - 2(4 \sin^2 x \cos^2 x)$
$= 1 - 8 \sin^2 x \cos^2 x$
$= R.H.S.$
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