How do I establish the identity?

cos^4x = 1/8(3+4cos(2x) + cos (4x))

Question

How do I establish the identity?

 cos^4x = 1/8(3+4cos(2x) + cos (4x))

solomonzelman · Answer

Use        $\large\color{black}{ \displaystyle \cos(2x)=2\cos^2x-1 }$

anonymous · Answer

ok, how would it work with having 1/8 and 4x

solomonzelman · Answer

lets just change the angles of (2x) and (4x) into angles of x.
(if that makes sense to you)

solomonzelman · Answer

what will 4cos(2x) be according to this identity ?

anonymous · Answer

4(2cos^2x-1) ?

solomonzelman · Answer

yes, and that expands/simplifies to ?

anonymous · Answer

8cos^2 x-4 ?

solomonzelman · Answer

yes, there you go

solomonzelman · Answer

$\large\color{black}{ \displaystyle \cos^4x = \frac{1}{8}(3+4\cos(2x) + \cos (4x))  }$

$\large\color{black}{ \displaystyle \cos^4x = \frac{1}{8}(3+4(2\cos^2x-1) + \cos (4x))  }$

$\large\color{black}{ \displaystyle \cos^4x = \frac{1}{8}(3+8\cos^2x-4 + \cos (4x))  }$

$\large\color{black}{ \displaystyle \cos^4x = \frac{1}{8}(8\cos^2x-1 + \cos (4x))  }$

so far see what is happening?

anonymous · Answer

sorry i have to leave, hopefully i can finish this up later