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## Verify the identity. cos 4x cos 2x = 2 – 2 sin^2 2x – 2 sin^2 x?

Hello,

a plus sign is missing at the left side:

cos(4x) + cos(2x) = 2 – 2sin²(2x) – 2sin²x

let’s work on the left side:

cos(4x) + cos(2x) =

let’s imagine the argument 4x as 2(2x):

cos[2(2x)] + cos(2x) =

let’s apply, to both cosines, the double-angle identity cos(2θ) = cos²θ – sin²θ:

[cos²(2x) – sin²(2x)] + (cos²x – sin²x) =

cos²(2x) – sin²(2x) + cos²x – sin²x =

let’s apply, to both cos²(2x) and cos²x, the identity cos²θ = 1 – sin²θ:

[1 – sin²(2x)] – sin²(2x) + (1 – sin²x) – sin²x =

1 – sin²(2x) – sin²(2x) + 1 – sin²x – sin²x =

ending with:

2 – 2sin²(2x) – 2sin²x (Q.E.D.)

I hope it’s helpful

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