Prove that
cos4x=8cos^4x-8cos^2x+1
so I'm guessing you begin with the left side cos4x=cos2(2x) then im kinda lost
You began the problem correctly. Continue to use the cos "double angle" formula, applying it twice.
cos4x = cos[2(2x)] = 2 cos^2(2x) - 1
= 2(2cos^2 x - 1)^2 -1
= 2*[4 cos^4 x -4cos^2 x +1) -1
= 8cos^2 x -8cos^2 x +1
4cos^{2}2x-1=cos4x
To prove the given equation cos4x = 8cos^4x - 8cos^2x + 1, we can start by using the double angle identity for cosine, which states:
cos(2θ) = 2cos^2(θ) - 1
Let's substitute 2θ with 2x:
cos4x = 2cos^2(2x) - 1
Now, we need to express cos^2(2x) in terms of cos^4(x) and cos^2(x) to continue simplifying. To do that, we can use the double angle identities again:
cos2x = 2cos^2(x) - 1
cos^2(x) = (1 + cos2x) / 2
We can rewrite the equation as:
cos4x = 2cos^2(2x) - 1
cos4x = 2(2cos^2(x) - 1)^2 - 1
Now, we expand and simplify:
cos4x = 2(4cos^4(x) - 4cos^2(x) + 1) - 1
cos4x = 8cos^4(x) - 8cos^2(x) + 2 - 1
cos4x = 8cos^4(x) - 8cos^2(x) + 1
Therefore, we have proven that cos4x = 8cos^4(x) - 8cos^2(x) + 1.