nathanralph
New member
- Joined
- Aug 27, 2005
- Messages
- 27
So I'm supposed to solve cos^4 2(theta) in terms of cosin with exponent 1. I have worked the problem out a few times with no luck. My steps are as follows:
cos^4 2(theta) = [cos^2 2(theta)]^2
= ([1+cos 4(theta)]/2)^2
= [1 + 2 cos 4(theta) + cos^2 4(theta)]/4
= 1/4 + 2/4 cos 4(theta) + 1/4 cos^2 4(theta)
= 1/4 + 1/2 cos 4(theta) + 1/4 [(1+ cos 8(theta)/2]
= 1/4 + 1/2 cos 4(theta) + 1/8 + 1/8 cos 8(theta)
= 3/8 + 1/2 cos 4(theta) + 1/8 cos 8(theta)
Sorry this is so long and tedious...just how it is solved. Am I calculating something wrong here? I've solved it several times with the same result, but am told I am incorrect.
cos^4 2(theta) = [cos^2 2(theta)]^2
= ([1+cos 4(theta)]/2)^2
= [1 + 2 cos 4(theta) + cos^2 4(theta)]/4
= 1/4 + 2/4 cos 4(theta) + 1/4 cos^2 4(theta)
= 1/4 + 1/2 cos 4(theta) + 1/4 [(1+ cos 8(theta)/2]
= 1/4 + 1/2 cos 4(theta) + 1/8 + 1/8 cos 8(theta)
= 3/8 + 1/2 cos 4(theta) + 1/8 cos 8(theta)
Sorry this is so long and tedious...just how it is solved. Am I calculating something wrong here? I've solved it several times with the same result, but am told I am incorrect.