Find an expression for \(cos(5\theta)\) as a fifth-degree polynomial in the variable \(cos\theta\).

AdamTaurus May 11, 2019

#1**+1 **

CPhil is probably going to give you the proper answer so I'll chime in with a method that's probably

above where you're currently at but is one of the more clever tricks in the toolbox

.\(\cos(5\theta)= Re(e^{i5\theta}) = \\ Re((e^{i\theta})^5) = Re((\cos(\theta)+i\sin(\theta))^5)\\ \text{Now apply the binomial expansion}\\ Re\left(\sum \limits_{k=0}^5\dbinom{5}{k}\cos^k(\theta)(i\sin(\theta))^{5-k}\right)\)

\(\text{We can see the only terms that are real will be }k=1,3,5\\ \text{and we end up with}\\ \cos(5\theta) = \cos^5(\theta)-10\cos^3(\theta)\sin^2(\theta)+5\cos(\theta)\sin^4(\theta)\)

\(\text{Oh, I see you need it entirely in }\cos(\theta)\text{. Well we just do this}\\ \cos(5\theta) = \cos^5(\theta)-10\cos^3(\theta)(1-\cos^2(\theta))+ 5\cos(\theta)(1-\cos^2(\theta))^2\)

You can do the final simplification.

.Rom May 11, 2019

#3**+1 **

I'll use x instead of theta

cos (5x) =

cos(3x + 2x) =

cos3xcos2x - sin3xsin2x =

cos(x +2x)cos2x - sin(x +2x)sin2x =

[cosxcos2x - sinxsin2x]cos2x - [sinxcos2x + sin2xcosx]sin2x =

cosxcos^2(2x) - sinxsin2xcos2x - sinxsin2xcos2x - cosxsin^2(2x) =

cosx [ cos^2(2x) - sin^2(2x] - 2sinx [ sin2xcos2x] =

cosx [cos^2(2x) - (1 - cos^2(2x) ] - 2sinx [ 2sinxcosx * (2cos^2x - 1) ] =

cosx [ 2cos^2(2x) - 1] - 2sinx [ 4sinxcos^3x - 2sinxcosx ] =

2cosx [cos(2x) ]^2 - cosx - 8sin^2xcos^3x + 4sin^2xcosx =

2cosx [ 2cos^2x - 1]^2 - cosx - 8[(1-cos^2x)cos^3x) ] + 4[(1-cos^2x )]cosx =

2cosx [ 4cos^4x - 4cos^2x + 1] - cosx - 8 [ cos^3x -cos^5x] + 4[ cosx - cos^3x ] =

8cos^5x - 8cos^3x + 2cosx - cosx - 8cos^3x + 8cos^5x + 4cosx - 4cos^3x =

16cos^5(x) - 20cos^3(x) + 5cos(x)

Someone should verify this....I could have made an error....!!!!

CPhill May 11, 2019