Cos^22x - sin^22x please helppppppp maths work
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Cos^22x - sin^22x please helppppppp maths work

[From: ] [author: ] [Date: 13-01-23] [Hit: ]
......
the answer is supposed to be cos4x

-
cos^2(2x) - sin^2(2x)

u = 2x

cos^2(u) - sin^2(u)

cos^2(x) - sin^2(x) = cos(2x)....

Therefore:
cos^2(u) - sin^2(u) = cos(2u)

Back substitute:
cos(2(2x)) = cos(4x)

-
solve(cos(2*x)^2 - sin(2*x)^2 = 0)

x = (1/8)*π or x = (3/8)*π
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