Exercise 1577Algebra › Complex numbersOlympiadAdded 3 Nov 2026

Trigonometric identities via complex numbers

#De Moivre#trigonometric identities#complex

Statement

Using De Moivre's formula, express $\sin(3\theta)$ in terms of $\sin\theta$ .

Hint

$\sin(3\theta)=\operatorname{Im}((\cos\theta+i\sin\theta)^3)$ .

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Expansion.
$(\cos\theta+i\sin\theta)^3=\cos^3\theta+3i\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta.$

Imaginary part.
$\sin(3\theta)=3\cos^2\theta\sin\theta-\sin^3\theta.$

Simplify. $\cos^2\theta=1-\sin^2\theta$ , so
$\sin(3\theta)=3\sin\theta(1-\sin^2\theta)-\sin^3\theta=3\sin\theta-4\sin^3\theta.$

Conclusion. $\sin(3\theta)=3\sin\theta-4\sin^3\theta$ .