How do I expand $\sin^5\theta$ using the complex exponential, in order to obtain:
$$\frac{1}{16}\sin 5\theta - \frac{5}{16}\sin 3\theta + \frac{5}{8}\sin\theta$$
Thank you.
$\endgroup$2 Answers
$\begingroup$Using this formula.
$$\exp({i\theta})^n = \exp(ni \theta)$$
$$(\cos(\theta) + i\sin(\theta))^n = \cos(n\theta) + i\sin(n\theta)$$
$$(\cos(\theta) + i\sin(\theta))^5 = \cos(5\theta) + i\sin(5\theta)$$
$$\cos^5(\theta) + 5i\cos^4(\theta)\sin(\theta) -10\cos^3(\theta)\sin^2(\theta) -10i \cos^2(\theta)\sin^3(\theta) + \cdots $$ $$5\cos(\theta)\sin^4(\theta) +i\sin^5(\theta) = \cos(5\theta) + i\sin(5\theta)$$
Equate imaginary parts:
$$5\cos^4(\theta)\sin(\theta)-10 \cos^2(\theta)\sin^3(\theta) + \sin^5(\theta) = \sin(5\theta)$$
Solve for $\sin^5(\theta)$:
$$\sin^5(\theta) = \sin(5\theta)-5\cos^4(\theta)\sin(\theta)+10 \cos^2(\theta)\sin^3(\theta)$$
Use trigonometric identities:
$$\sin^5(\theta) = \dfrac{1}{16} (\sin(5 \theta)-5 \sin(3 \theta) + 10 \sin(\theta))$$
$\endgroup$ 4 $\begingroup$Note that $$\sin(\theta) = {e^{i\theta} - e^{-i\theta}\over 2i}.$$ Use the binomial theorem.
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