Express cos^(4)2x in terms of cosine with an exponent of 1?

Question

Hi, this is my first time using this site.

So I have been staring at this trigonometric equation for a long while and looking for what it asks for. For what I know, I think I have to use sum of difference identities, but I am not so sure where to begin from there. I know I am working for an answer of cosine, but then I am lost as to what identity I should use, or if I am using the right identity.

I really appreciate any help for this problem and I thank you.

Answer 1

You will need to use the power reducing identity for cosine:

   cos²x = (1+cos(2x))/2

   cos²(2x) = (1+cos(4x))/2

   cos⁴(2x)= [(1+cos(4x))/2]²

 

 

Here is our first bout of simplification:

   cos⁴(2x) = [(1+cos(4x))/2]²

   cos⁴(2x) = (1/4)*[1 + cos(4x)]²

   cos⁴(2x) = (1/4)*(1 + 2cos(4x) + cos²(4x))

 

We are almost there! Use the power reducing identity one more time:

   cos⁴(2x) = (1/4)*[1 + 2cos(4x) + (1 + cos(2x))/2]

 

Here is our second (and final) bout of simplification:

   cos⁴(2x) = (1/4)*[1 + 2cos(4x) + (1 + cos(2x))/2]

   cos⁴(2x) = (1/4)*[1 + 2cos(4x) + (1/2)*(1 + cos(2x))]

   cos⁴(2x) = (1/8)*[2 + 4cos(4x) + 1 + cos(2x)]                <-- Here I factored out a (1/2)

   cos⁴(2x) = (1/8)*[3 + cos(2x) + 4cos(4x)]                      <-- Combine like terms and rearrange

 

There are several ways that you can rewrite this depending on your teacher's preferences. This one is perfectly valid. I hope this helps.