prove cos 6 x = 1/32(cos 6x + 6 Cos 4x + 15 Cos 2x + 10)

Question

Arun · Answer

I have solved it with a simple and smart approach.

cos^6x=((e^ix+e^-ix)/2)^6
=1/2^6(e^6ix+6e^5ixe^-ix+15e^4ixe^-2ix
+20e^3ixe^-3ix+15e^2ixe^-4ix
+6e^ixe^-5ix+e^-6ix)
=1/32(((e^6ix+e^-6ix)/2)
+6((e^4ix+e^-4ix)/2)
+15((e^2ix+e^-2ix)/2)
+10)
cos^6x=1/32(15cos2x
+6cos4x
+cos6x
+10)

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prove cos 6 x = 1/32(cos 6x + 6 Cos 4x + 15 Cos 2x + 10)

prove cos^{6_{x = 1/32(cos 6x + 6 Cos 4x + 15 Cos 2x + 10)}}

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prove cos 6 x = 1/32(cos 6x + 6 Cos 4x + 15 Cos 2x + 10)

prove cos6x = 1/32(cos 6x + 6 Cos 4x + 15 Cos 2x + 10)

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prove cos^{6_{x = 1/32(cos 6x + 6 Cos 4x + 15 Cos 2x + 10)}}