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