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Verify the identity. cos 4x + cos 2x = 2 - 2 sin^2 2x - 2 sin^2 x

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cos(4x)=cos(2x+2x) expand that^
\(\cos4x=1-2\sin^2 2x\) \(\cos2x=1-2\sin^2x\) does that help?
It did, but could you please show me how cos 4x = 1 - 2 sin^2 2x?

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Other answers:

ya sure. \(\cos4x=\cos(2x+2x)\) \(=\cos2x\cos2x-\sin2x\sin2x\) \(=\cos^2 2x-\sin^2 2x\) \(=1-\sin^2 2x-\sin^2 2x\) \((\cos^2 2x=1-\sin^2 2x)\) \(=1-2\sin^2 2x\)
Do you have a reference where it says cos^2 2x = 1−sin^2 2x? I think it would be useful for me since I have a formula sheet but I don't see that on here...
\(\cos^2x+\sin^2x=1\) Similarly \(\cos^2 2x+\sin^2 2x=1\) \(\cos^2 2x=1-\sin^2 2x\)
Thank you for your time and assistance. :)

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