Sine to Power of Even Integer
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Theorem
\(\ds \sin^{2 n} \theta\) | \(=\) | \(\ds \frac 1 {2^{2 n} } \binom {2 n} n + \frac {\paren {-1}^n} {2^{2 n - 1} } \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k \binom {2 n} k \map \cos {2 n - 2 k} \theta\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2^{2 n} } \binom {2 n} n + \frac {\paren {-1}^n} {2^{2 n - 1} } \paren {\cos 2 n \theta - \binom {2 n} 1 \map \cos {2 n - 2} \theta + \cdots + \paren {-1}^{n - 1} \binom {2 n} {n - 1} \cos 2 \theta}\) |
Proof
First, by Euler's Sine Identity 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:
\(\ds \paren {e^{i \theta} - e^{-i \theta} }^{2 n}\) | \(=\) | \(\ds \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k e^{i \paren {2 n - k} \theta} e^{-i k \theta}\) | Binomial Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k e^{i \paren {2 n - 2 k}\theta}\) | Power of Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \cos {2 n - 2 k} \theta + i \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \sin {2 n - 2 k} \theta\) | Euler's Formula |
So we have:
- $\ds \sin^{2 n} \theta = \dfrac {\paren {-1}^n} {2^{2 n} } \paren {\sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \cos {2 n - 2 k} \theta + i \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \sin {2 n - 2 k} \theta}$
Now we look at each of the terms in the parentheses:
\(\ds \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \map \sin {2 n - 2 k} \theta\) | \(=\) | \(\ds \paren {-1}^n \sum_{k \mathop = -n}^n {2 n \choose n + k} \paren {-1}^k \map \sin {-2 k \theta}\) | replacing $k \mapsto k + n$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^n \sum_{k \mathop = -n}^{-1} {2 n \choose n + k} \paren {-1}^k \map \sin {-2 k \theta} + \paren {-1}^n \sum_{k \mathop = 1}^n {2 n \choose n + k} \paren {-1}^k \map \sin {-2 k \theta}\) | Zeroes of Sine and Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n - k} \sin 2 k \theta + {2 n \choose n + k} \map \sin {-2 k \theta} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n + k} \sin 2 k \theta + {2 n \choose n + k} \map \sin {-2 k \theta} }\) | Symmetry Rule for Binomial Coefficients | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n + k} \sin 2 k \theta - {2 n \choose n + k} \sin 2 k \theta}\) | Sine Function is Odd | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
We find, for the remaining term:
\(\ds \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \cos {2 n - 2 k} \theta\) | \(=\) | \(\ds \paren {-1}^n \sum_{k \mathop = -n}^n {2 n \choose n + k} \paren {-1}^k \map \cos {-2 k \theta}\) | replacing $k \mapsto k + n$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^n \sum_{k \mathop = -n}^{-1} {2 n \choose n + k} \paren {-1}^k \map \cos {-2 k \theta} + \paren {-1}^n {2 n \choose n} + \paren {-1}^n \sum_{k \mathop = 1}^n {2 n \choose n + k} \paren {-1}^k \map \cos {-2 k \theta}\) | Cosine of Zero is One | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^n {2 n \choose n} + \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n - k} \cos 2 k \theta + {2 n \choose n + k} \map \cos {-2 k \theta} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^n {2 n \choose n} + \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n - k} \cos 2 k \theta + {2 n \choose n + k} \cos 2 k \theta}\) | Cosine Function is Even | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^n {2 n \choose n} + \paren {-1}^n 2 \sum_{k \mathop = 1}^n \paren {-1}^k {2 n \choose n - k} \cos 2 k \theta\) | Symmetry Rule for Binomial Coefficients |
Thus we have:
\(\ds \sin^{2 n} \theta\) | \(=\) | \(\ds \frac {\paren {-1}^n} {2^{2 n} } \paren {\paren {-1}^n {2 n \choose n} + \paren {-1}^n 2 \sum_{k \mathop = 1}^n \paren {-1}^k {2 n \choose n - k} \cos 2 k \theta}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2^{2 n} } {2 n \choose n} + \frac 1 {2^{2 n - 1} } \sum_{k \mathop = 1}^n \paren {-1}^k {2 n \choose n - k} \cos 2 k \theta\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2^{2 n} } {2 n \choose n} + \frac {\paren {-1}^n} {2^{2 n - 1} } \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k {2 n \choose k} \map \cos {2 n - 2 k} \theta\) | replacing $k \mapsto n-k$ |
as required.
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.72$
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $130 \ \text{(b)}$