Quadruple Angle Formulas/Sine

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Theorem

$\sin 4 \theta = 4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta$

where $\sin$ and $\cos$ denote sine and cosine respectively.


Corollary 1

For all $\theta$ such that $\theta \ne 0, \pm \pi, \pm 2 \pi \ldots$

$\dfrac {\sin 4 \theta} {\sin \theta} = 8 \cos^3 \theta - 4 \cos \theta$

where $\sin$ denotes sine and $\cos$ denotes cosine.


Corollary 2

For all $\theta$ such that $\theta \ne 0, \pm \pi, \pm 2 \pi \ldots$

$\dfrac {\sin 4 \theta} {\sin \theta} = 2 \cos 3 \theta + 2 \cos \theta$

where $\sin$ denotes sine and $\cos$ denotes cosine.


Proof 1

\(\displaystyle \sin \left({4 \theta}\right)\) \(=\) \(\displaystyle \sin \left({3 \theta + \theta}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \sin 3 \theta \cos \theta + \cos 3 \theta \sin \theta\) Sine of Sum
\(\displaystyle \) \(=\) \(\displaystyle \left({3 \sin \theta - 4 \sin^3 \theta}\right) \cos \theta + \left({4 \cos^3 \theta - 3 \cos \theta}\right) \sin \theta\) Triple Angle Formula for Sine and Triple Angle Formula for Cosine
\(\displaystyle \) \(=\) \(\displaystyle 3 \sin \theta \cos \theta - 4 \sin^3 \theta \cos \theta + 4 \cos^3 \theta \sin \theta - 3 \cos \theta \sin \theta\)
\(\displaystyle \) \(=\) \(\displaystyle 3 \sin \theta \cos \theta - 4 \sin^3 \theta \cos \theta + 4 \cos \theta \left({1 - \sin^2 \theta}\right) \sin \theta - 3 \cos \theta \sin \theta\) Sum of Squares of Sine and Cosine
\(\displaystyle \) \(=\) \(\displaystyle 4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta\) multiplying out and gathering terms

$\blacksquare$


Proof 2

We have:

\(\displaystyle \cos 4 \theta + i \sin 4 \theta\) \(=\) \(\displaystyle \paren {\cos \theta + i \sin \theta}^4\) De Moivre's Formula
\(\displaystyle \) \(=\) \(\displaystyle \paren {\cos \theta}^4 + \binom 4 1 \paren {\cos \theta}^3 \paren {i \sin \theta} + \binom 4 2 \paren {\cos \theta}^2 \paren {i \sin \theta}^2\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \binom 4 3 \paren {\cos \theta} \paren {i \sin \theta}^3 + \paren {i \sin \theta}^4\) Binomial Theorem
\(\displaystyle \) \(=\) \(\displaystyle \cos^4 \theta + 4 i \cos^3 \theta \sin \theta - 6 \cos^2 \theta \sin^2 \theta\) substituting for binomial coefficients
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle 4 i \cos \theta \sin^3 \theta + \sin^4 \theta\) and using $i^2 = -1$
\(\text {(1)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle \cos^4 \theta - 6 \cos^2 \theta \sin^2 \theta + \sin^4 \theta\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle 4 i \cos^3 \theta \sin \theta - 4 i \cos \theta \sin^3 \theta\) rearranging


Hence:

\(\displaystyle \sin 4 \theta\) \(=\) \(\displaystyle 4 \cos^3 \theta \sin \theta - 4 \cos \theta \sin^3 \theta\) equating imaginary parts in $(1)$
\(\displaystyle \) \(=\) \(\displaystyle 4 \cos \theta \paren {1 - \sin^2 \theta} \sin \theta - 4 \cos \theta \sin^3 \theta\) Sum of Squares of Sine and Cosine
\(\displaystyle \) \(=\) \(\displaystyle 4 \cos \theta \sin \theta - 8 \sin^3 \theta \cos \theta\) multiplying out and gathering terms

$\blacksquare$


Sources