1
$\begingroup$

I'm looking at the solution manual and I have no idea how they convert this.

$ k \cos {3\theta} = k [4\cos^{3} {\theta} - 3 \cos{\theta}] = k[\alpha P_{3}(\cos\theta) + \beta P_{1}\cos{\theta}] $

(I know that any 3rd order polynomial can be express as a linear combination of the first for Legendre polynomials; in this case, since the polynomial is odd, I only need P1 and P3)

  • 0
    It appears that $1 \to 2$ is [here at this link](http://answers.yahoo.com/question/index?qid=20080617083026AAgX3vQ).2012-03-24

1 Answers 1

1

You don't need to upvote, but for clarity here is how from $1$ to $2$

\begin{align*} cos(3\theta) &= cos(2\theta+\theta)\\ &= cos(2\theta)cos(\theta) - sin(2\theta)sin(\theta)\\ &= \left[2cos²(\theta)-1\right]cos(\theta) - \left[2sin(\theta)cos(\theta)\right]sin(\theta) \\ &= 2cos³(\theta) - cos(\theta) - 2sin²(\theta)cos(\theta) \\ &= 2cos³(\theta) - cos(\theta) - 2\left[1 - cos²(\theta)\right]cos(\theta)\\ &= 2cos³(\theta) - cos(\theta) - 2cos(\theta)\left[1 - cos²(\theta)\right]\\ &= 2cos³(\theta) - cos(\theta) - 2cos(\theta) + 2cos³(\theta)\\ &= 4cos³(\theta) - 3cos(\theta) \end{align*}

  • 0
    A friendly $\TeX$ pointer: Using `$cos$` looks a bit uglier than using `$\cos$` which looks like $\cos$. (I'll +1 this!) Regards,2012-03-24