Prove that: $\cos 6 x=32 x \cos ^{6} x-48 \cos ^{4} x+18 \cos ^{2} x-1$
Prove that: $\cos 6 x=32 x \cos ^{6} x-48 \cos ^{4} x+18 \cos ^{2} x-1$
$L.H.S.$ $=\cos 6 x$
$=\cos 3(2 x)$
$=4 \cos ^{3} 2 x-3 \cos 2 x\left[\cos 3 A=4 \cos ^{3} A-3 \cos A\right]$
$=4\left[\left(2 \cos ^{2} x-1\right)^{3}-3\left(2 \cos ^{2} x-1\right)\right]\left[\cos 2 x=2 \cos ^{2} x-1\right]$
$=4\left[\left(2 \cos ^{2} x\right)^{3}-(1)^{3}-3\left(2 \cos ^{2} x\right)^{2}+3\left(2 \cos ^{2} x\right)\right]-6 \cos ^{2} x+3$
$=4\left[8 \cos ^{6} x-1-12 \cos ^{4} x+6 \cos ^{2} x\right]-6 \cos ^{2} x+3$
$=32 \cos ^{6} x-4-48 \cos ^{4} x+24 \cos ^{2} x-6 \cos ^{2} x+3$
$=32 \cos ^{6} x-48 \cos ^{4} x+18 \cos ^{2} x-1$
$=\operatorname{R.H.S}$
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