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Prove the following cos6x = 32 cos^6x - 48cos^4x +18 cos^2x -1

Q(25)  Prove the following

\small \cos6x = 32\cos^{6}x -48\cos^{4}x + 18\cos^{2}x-1

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We know that
 cos 3x = 4\cos^{3}x - 3cos x
we use this in our problem
we can write cos 6x as cos 3(2x)
        cos 3(2x) = 4\cos^{3}2x - 3 cos 2x
                         =  4(2\cos^{2}x - 1)^{3}  - 3(2\cos^{2}x - 1)                                                                                        (\because \cos 2x = 2\cos^{2}x - 1)
                         = 4[(2cos^{2}x)^{3} -(1)^{3} -3(2cos^{2}x)^{2}(1) + 3(2cos^{2}x)(1)^{2}]    -6\cos^{2}x + 3             (\because (a-b)^{3} = a^{3} - b^{3} - 3a^{2}b+ 3ab^{2})
                         = 32cos^{6}x - 4 - 48cos^{4}x + 24cos^{2}x  - 6\cos^{2}x + 3
                         =   32cos^{6}x  - 48cos^{4}x + 18cos^{2}x - 1  = R.H.S.

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