By using the identity cos2A=1-2sin^2A show that cos4x = 1-8sin^2xcos^2x
cos 4x = 1 - 2sin^2 (2x)
= 1 - 2(sin 2x)(sin 2x)
= 1 - 2(2sinxcosx)(2sinxcosx)
= 1 - 8(sinxcosx)^2
or
= 1 - 8(sin^2 x)(cos^2 x) as required
thank you!
To prove that cos4x = 1-8sin^2xcos^2x using the identity cos2A=1-2sin^2A, you need to apply the double angle formula repeatedly.
First, let's express cos4x using the double angle formula for cos2A:
cos4x = cos(2*2x)
Using the double angle formula, this can be written as:
cos4x = cos^2(2x) - sin^2(2x)
Next, substitute 2x for A in the identity cos2A=1-2sin^2A:
cos^2(2x) - sin^2(2x) = 1 - 2sin^2(2x)
Now, writing sin^2(2x) as (2sinxcosx)^2, we can simplify the equation further:
cos^2(2x) - sin^2(2x) = 1 - 2(2sinxcosx)^2
Expanding (2sinxcosx)^2 gives:
cos^2(2x) - sin^2(2x) = 1 - 8sin^2xcos^2x
Therefore, we have shown that cos4x = 1 - 8sin^2xcos^2x using the identity cos2A = 1 - 2sin^2A.