# Sine to Power of Even Integer

## Theorem

 $\displaystyle \sin^{2 n} \theta$ $=$ $\displaystyle \frac 1 {2^{2 n} } \binom {2 n} n + \frac {\paren {-1}^n} {2^{2 n - 1} } \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k \binom {2 n} k \map \cos {2 n - 2 k} \theta$ $\displaystyle$ $=$ $\displaystyle \frac 1 {2^{2 n} } \binom {2 n} n + \frac {\paren {-1}^n} {2^{2 n - 1} } \paren {\cos 2 n \theta - \binom {2 n} 1 \map \cos {2 n - 2} \theta + \cdots + \paren {-1}^{n - 1} \binom {2 n} {n - 1} \cos 2 \theta}$

## Proof

First, by Sine Exponential Formulation we have:

$\sin \theta = \dfrac 1 {2 i} \paren {e^{i \theta} - e^{-i \theta} }$

Therefore by Power of Product:

$\sin^{2 n} \theta = \dfrac 1 {\paren {2 i}^{2 n} } \paren {e^{i \theta} - e^{-i \theta} }^{2 n}$

Now by Power of Product and the Power of Power:

$\dfrac 1 {\paren {2 i}^{2 n} } = \dfrac 1 {2^{2 n} \paren {-1}^n} = \dfrac {\paren {-1}^n} {2^{2 n} }$

Thus:

$\sin^{2 n} \theta = \dfrac {\paren {-1}^n} {2^{2 n} } \paren {e^{i \theta} - e^{-i \theta} }^{2 n}$

Now:

 $\displaystyle \paren {e^{i \theta} - e^{-i \theta} }^{2 n}$ $=$ $\displaystyle \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k e^{i \paren {2 n - k} \theta} e^{-i k \theta}$ Binomial Theorem $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k e^{i \paren {2 n - 2 k}\theta}$ Power of Product $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \cos {2 n - 2 k} \theta + i \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \sin {2 n - 2 k} \theta$ Euler's Formula

So we have:

$\displaystyle \sin^{2 n} \theta = \dfrac {\paren {-1}^n} {2^{2 n} } \paren {\sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \cos {2 n - 2 k} \theta + i \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \sin {2 n - 2 k} \theta}$

Now we look at each of the terms in the parentheses:

 $\displaystyle \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \map \sin {2 n - 2 k} \theta$ $=$ $\displaystyle \paren {-1}^n \sum_{k \mathop = -n}^n {2 n \choose n + k} \paren {-1}^k \map \sin {-2 k \theta}$ replacing $k \mapsto k + n$ $\displaystyle$ $=$ $\displaystyle \paren {-1}^n \sum_{k \mathop = -n}^{-1} {2 n \choose n + k} \paren {-1}^k \map \sin {-2 k \theta} + \paren {-1}^n \sum_{k \mathop = 1}^n {2 n \choose n + k} \paren {-1}^k \map \sin {-2 k \theta}$ Zeroes of Sine and Cosine $\displaystyle$ $=$ $\displaystyle \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n - k} \sin 2 k \theta + {2 n \choose n + k} \map \sin {-2 k \theta} }$ $\displaystyle$ $=$ $\displaystyle \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n + k} \sin 2 k \theta + {2 n \choose n + k} \map \sin {-2 k \theta} }$ Symmetry Rule for Binomial Coefficients $\displaystyle$ $=$ $\displaystyle \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n + k} \sin 2 k \theta - {2 n \choose n + k} \sin 2 k \theta}$ Sine Function is Odd $\displaystyle$ $=$ $\displaystyle 0$

We find, for the remaining term:

 $\displaystyle \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \cos {2 n - 2 k} \theta$ $=$ $\displaystyle \paren {-1}^n \sum_{k \mathop = -n}^n {2 n \choose n + k} \paren {-1}^k \map \cos {-2 k \theta}$ replacing $k \mapsto k + n$ $\displaystyle$ $=$ $\displaystyle \paren {-1}^n \sum_{k \mathop = -n}^{-1} {2 n \choose n + k} \paren {-1}^k \map \cos {-2 k \theta} + \paren {-1}^n {2 n \choose n} + \paren {-1}^n \sum_{k \mathop = 1}^n {2 n \choose n + k} \paren {-1}^k \map \cos {-2 k \theta}$ Cosine of Zero is One $\displaystyle$ $=$ $\displaystyle \paren {-1}^n {2 n \choose n} + \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n - k} \cos 2 k \theta + {2 n \choose n + k} \map \cos {-2 k \theta} }$ $\displaystyle$ $=$ $\displaystyle \paren {-1}^n {2 n \choose n} + \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n - k} \cos 2 k \theta + {2 n \choose n + k} \cos 2 k \theta}$ Cosine Function is Even $\displaystyle$ $=$ $\displaystyle \paren {-1}^n {2 n \choose n} + \paren {-1}^n 2 \sum_{k \mathop = 1}^n \paren {-1}^k {2 n \choose n - k} \cos 2 k \theta$ Symmetry Rule for Binomial Coefficients

Thus we have:

 $\displaystyle \sin^{2 n} \theta$ $=$ $\displaystyle \frac {\paren {-1}^n} {2^{2 n} } \paren {\paren {-1}^n {2 n \choose n} + \paren {-1}^n 2 \sum_{k \mathop = 1}^n \paren {-1}^k {2 n \choose n - k} \cos 2 k \theta}$ $\displaystyle$ $=$ $\displaystyle \frac 1 {2^{2 n} } {2 n \choose n} + \frac 1 {2^{2 n - 1} } \sum_{k \mathop = 1}^n \paren {-1}^k {2 n \choose n - k} \cos 2 k \theta$ $\displaystyle$ $=$ $\displaystyle \frac 1 {2^{2 n} } {2 n \choose n} + \frac {\paren {-1}^n} {2^{2 n - 1} } \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k {2 n \choose k} \map \cos {2 n - 2 k} \theta$ replacing $k \mapsto n-k$

as required.

$\blacksquare$