Verified Answer :

Julik

\(\\ sin^2 \alpha +cos^2 \alpha =1 \\ cos2 \alpha =1-2sin^2 \alpha \\cos( \alpha + \beta )=cos \alpha cos \beta -sin \alpha sin \beta \\ \\ cos4x=cos(2x+2x)=cos2xcos2x-sin2xsin2x = \\ =cos^22x-sin^22x \\ cos^22x=1-sin^22x \\ cos^22x-sin^22x=1-sin^22x-sin^22x=1-2sin^22x \\ cos4x=1-2sin^22x \\ cos2x =1-2sin^2x \\ \\ cos4x+cos2x=1-2sin^22x+1-2sin^2x= \\ =2-2sin^22x-2sin^2x \\ \\ cos4x+cos2x=2-2sin^22x-2sin^2x\)

taskmasters

By using trigonometric identities, we can solve the identity given above. We just need to work on one side and make it equal to the other side. We choose the right side:

2−2sin2(2x)−2sin2(x)=1+1−2sin2(2x)−2sin2(x) = (1−2sin2(2x))+(1−2sin2(x)) + (1−2sin2(2x))+(1−2sin2(x)) =cos(4x)+cos(2x)

cos 4x + cos 2x = cos 4x + cos 2x

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