# Use de Moivre's formula to write $\cos(n\theta)$ as a polynomial of $\cos\theta$ and $\sin\theta$

Use the binomial formula and de Moivre’s formula to write $$\cos n\theta + i \sin n\theta = \sum_{k=0}^n \binom{n}{k} \cos^{n-k} \theta (i \sin \theta)^k \quad (n = 0, 1, 2, \dotsc).$$ Then define the integer $m$ by means of the equations $$m = \begin{cases} n/2 & \text{if n is even}, \\ (n-1)/2 & \text{if n is odd} \end{cases}$$ and use the above summation to show that $$\cos n\theta = \sum_{k=0}^m \binom{n}{2k} (-1)^k \cos^{n-2k} \theta \sin^{2k} \theta \quad (n = 0, 1, 2, \dotsc).$$

I can use the binomial formula and de Moivre's formula to get the first equation but after I define $m$ as above, I don't know how to change $k$ to $2k$ and get the formula.

• You need to cancel $isin\theta$ with $-isin\theta$ how would you do it? Dec 31 '15 at 5:28
• $\cos(n\theta)$ is the real part of the sum in the first equation. The real part is given by the sum over the even $k$. For such $k$, let $k=2j$. Write down the real part as a sum over $j$. Then change the dummy summation variable to $k$. They really should not have used $k$ for the last expression, it just confuses people. Dec 31 '15 at 5:29
• it really helps,thx a lot! Dec 31 '15 at 5:41

$$\cos n\theta + i\sin n\theta=\sum_{k=0}^n{n \choose k}\cos^{n-k}\theta(i\sin\theta)^k\tag{1}$$ $$\cos (-n\theta) + i\sin (-n\theta)=\sum_{k=0}^n{n \choose k}\cos^{n-k}(-\theta)(i\sin(-\theta))^k$$ $$\cos n\theta - i\sin n\theta=\sum_{k=0}^n{n \choose k}\cos^{n-k}\theta\cdot(-1)^k(i\sin\theta)^k\tag{2}$$ $\dfrac{(1)+(2)}2$: $$\cos n\theta=\sum_{k=even}^n{n \choose k}\cos^{n-k}\theta(i\sin\theta)^k=\sum_{k=0}^{\lfloor \frac n2 \rfloor}{n \choose 2k}\cos^{n-2k}\theta(i\sin\theta)^{2k}\\=\sum_{k=0}^{\lfloor \frac n2 \rfloor}{n \choose 2k}\cos^{n-2k}\theta\cdot i^{2k}\sin^{2k}\theta=\sum_{k=0}^{\lfloor \frac n2 \rfloor}{n \choose 2k}\cos^{n-2k}\theta\cdot (-1)^{k}\sin^{2k}\theta$$