Question about proof of general formula for $\sin^n \theta$

Question

I was reading the wikipedia page about list of trigonometric identites and came across the following identities

for $$\cos^n \theta $$ when n is odd:-

$$\cos^n \theta = \frac{2}{2^n}\sum_{k=0}^{\frac{n-1}{2}} {n\choose k}\cos ((n-2k)\theta)$$

for $$\cos^n \theta $$ when n is even:-

$$\cos^n \theta = \frac{1}{2^n}{n\choose \frac{n}{2}} + \frac{2}{2^n}\sum_{k=0}^{\frac{n}{2} - 1}{n\choose k}\cos((n-2k)\theta)$$

I tried to prove it using the binomial theorem expansion of $(\cos \theta + i \cdot \sin \theta)^n$ and isolating the $\cos^n \theta$ term however i do not understand how to proceed any help would be appreciated

Answer 1

Remember that: $$\cos x=\frac{e^{ix}+e^{-ix}}{2}$$ Using the binomial theorem we obtain that: $$\begin{align}\cos^nx&=\frac{1}{2^n}\sum_{k=0}^n\binom{n}{k}\left(e^{ix}\right)^{n-k}\left(e^{-ix}\right)^{k}\\&=\frac{1}{2^n}\sum_{k=0}^n\binom{n}{k}e^{i(n-2k)x}\\&=\frac{1}{2^n}\sum_{k=0}^n\binom{n}{k}\left[\cos((n-2k)x)+i\sin((n-2k)x)\right]\end{align}$$ Now, since we know that the result $\cos^nx$ must be real, the sum of the imaginary terms in the RHS of the previous equation cancels out. So we can ignore it, concluding that: $$\boxed{\cos^nx=\frac{1}{2^n}\sum_{k=0}^n\binom{n}{k}\cos((n-2k)x)}$$ If $n$ is odd we have that: $$\begin{align}\cos^nx&=\frac{1}{2^n}\sum_{k=0}^n\binom{n}{k}\cos((n-2k)x)\\&=\frac{1}{2^n}\sum_{k=0}^{\frac{n-1}{2}}\binom{n}{k}\cos((n-2k)x)+\frac{1}{2^n}\sum_{k=\frac{n-1}{2}+1}^n\binom{n}{k}\cos((n-2k)x)\\&=\color{red}\ast\end{align}$$ If $\ell:=k-\frac{n+1}{2}$ then, since $k$ goes from $\frac{n+1}{2}$ to $n$, $\ell$ goes from $0$ to $\frac{n-1}{2}$. Using the binomial property $\binom{a}{b}=\binom{a}{a-b}$ we have that: $$\begin{align}\color{red}\ast&=\frac{1}{2^n}\sum_{k=0}^{\frac{n-1}{2}}\binom{n}{k}\cos((n-2k)x)+\frac{1}{2^n}\sum_{\ell=0}^\frac{n-1}{2}\binom{n}{\frac{n+1}{2}+\ell}\cos\left(\left(n-2\left(\frac{n+1}{2}+\ell\right)\right)x\right)\\&=\frac{1}{2^n}\sum_{k=0}^{\frac{n-1}{2}}\binom{n}{k}\cos((n-2k)x)+\frac{1}{2^n}\sum_{\ell=0}^\frac{n-1}{2}\binom{n}{n-\frac{n+1}{2}-\ell}\cos\left((-2\ell-1)x\right)\\&=\frac{1}{2^n}\sum_{k=0}^{\frac{n-1}{2}}\binom{n}{k}\cos((n-2k)x)+\frac{1}{2^n}\sum_{\ell=0}^\frac{n-1}{2}\binom{n}{\frac{n-1}{2}-\ell}\cos\left((2\ell+1)x\right)\\&=\color{blue}\blacktriangle\end{align}$$ Let's change variable $m=\frac{n-1}{2}-\ell$. Then: $$\begin{align}\color{blue}\blacktriangle&=\frac{1}{2^n}\sum_{k=0}^{\frac{n-1}{2}}\binom{n}{k}\cos((n-2k)x)+\frac{1}{2^n}\sum_{m=0}^\frac{n-1}{2}\binom{n}{m}\cos\left(\left(2\left(\frac{n-1}{2}-m\right)+1\right)x\right)\\&=\frac{1}{2^n}\sum_{k=0}^{\frac{n-1}{2}}\binom{n}{k}\cos((n-2k)x)+\frac{1}{2^n}\sum_{m=0}^\frac{n-1}{2}\binom{n}{m}\cos((n-2m)x)\end{align}$$ Thus, since the role of $k$ and $m$ is the same, we finally conclude that: $$\boxed{\cos^nx=\frac{2}{2^n}\sum_{k=0}^{\frac{n-1}{2}}\binom{n}{k}\cos((n-2k)x)}$$ as we wanted. Proceeding in a similar manner we can prove the formula when $n$ is even.

Answer 2

$$ \begin{aligned} \cos ^n \theta & =\left(\frac{e^{\theta i}+e^{-\theta i}}{2}\right)^n \\ & =\frac{1}{2^n} \sum_{k=0}^n\left(\begin{array}{l} n \\ k \end{array}\right) e^{(n-k) \theta i} \cdot e^{-k\theta i} \\ & =\frac{1}{2^n} \sum_{k=0}^n\left(\begin{array}{l} n \\ k \end{array}\right) e^{(n-2 k) \theta i} \end{aligned} $$ When $n$ is odd, $$ \begin{aligned} \cos ^n \theta&=\frac{1}{2^n} \sum_{k=0}^{\frac {n-1}{2}}\left(\begin{array}{l} n \\ k \end{array}\right)\left(e^{(n-2 k) \theta i}+e^{-(n-2k) \theta i}\right) \\ & =\frac{1}{2^n} \sum_{k=0}^{\frac {n-1}{2}}\left(\begin{array}{l} n \\ k \end{array}\right) \cdot 2 \cos(n-2 k) \theta \\ & =\frac{2}{2^n} \sum_{k=0}^{\frac {n-1}{2}}\left(\begin{array}{l} n \\ k \end{array}\right) \cos((n-2 k)) \theta \\ & \end{aligned} $$ When $n$ is even, $$ \begin{gathered} \cos ^n \theta=\frac{1}{2^n}\left[\left(\begin{array}{l} n \\ \frac{n}{2} \end{array}\right)+\sum_{k=0}^{\frac{n}{2}-1}\left(\begin{array}{l} n \\ k \end{array}\right)\left(e^{(n-2 k) \theta i}+e^{-(n-2 k) \theta i}\right)\right] \\ =\frac{1}{2^n}\left(\begin{array}{l} n \\ \frac{n}{2} \end{array}\right)+\frac{2}{2^n} \sum_{k=0}^{\frac{n}{2}-1}\left(\begin{array}{l} n \\ k \end{array}\right) (\cos (n-2 k) \theta ) \end{gathered} $$

---

Formula for sine, replacing $\theta $ by $\frac{\pi}{2}-\theta$, we have for any odd integer $n$, $$ \sin ^n \theta=\frac{2}{2^n} \sum_{k=0}^{\frac{n-1}{2}}\left(\begin{array}{l} n \\ k \end{array}\right) \cos \left((n-2 k)\left(\frac{\pi}{2}-\theta\right)\right) $$

For any even integer $n$, we have

$$ \sin ^n \theta=\frac{1}{2^n}\left(\begin{array}{c} n \\ \frac{n}{2} \end{array}\right)+\frac{2}{2^n} \sum_{k=0}^{\frac{n}{2}-1}\left(\begin{array}{l} n \\ k \end{array}\right) \cos \left((n-2 k)\left(\frac{\pi}{2}-\theta\right)\right) $$