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Prove that
$\cos 6 x=32 \cos ^6 x-48 \cos ^4 x+18 \cos ^2 x-1$
$\cos 6 x=32 \cos ^6 x-48 \cos ^4 x+18 \cos ^2 x-1$
Solution:
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Verified Answer
$\cos 6 x=\cos 3(2 x)=4 \cos ^3 2 x-3 \cos 2 x$
Put $\cos 2 x=2 \cos ^2 x-1$
$\cos 6 x=4\left(2 \cos ^2 x-1\right)^3-3\left(2 \cos ^2 x-1\right)$
$=4\left(8 \cos ^6 x-12 \cos ^4 x+6 \cos ^2 x-1\right)-6 \cos ^2 x+3$
$=32 \cos ^6 x-48 \cos ^4 x+18 \cos ^2 x-1=$ RHS
Put $\cos 2 x=2 \cos ^2 x-1$
$\cos 6 x=4\left(2 \cos ^2 x-1\right)^3-3\left(2 \cos ^2 x-1\right)$
$=4\left(8 \cos ^6 x-12 \cos ^4 x+6 \cos ^2 x-1\right)-6 \cos ^2 x+3$
$=32 \cos ^6 x-48 \cos ^4 x+18 \cos ^2 x-1=$ RHS
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