Use De Moivre to derive the sine, cosine and tangent triple-angle formulae.

01 / Standard#Your turn

Official paper · jm02-2023 · 4(b)(i) · PDF 6

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Expand (cosθ+i sinθ)³.

Hint 2

Compare real and imaginary parts, then divide where defined.

Worked solution

Equate the expansion to cis(3θ).
$\cos3\theta+i\sin3\theta=(\cos^3\theta-3\cos\theta\sin^2\theta)+i(3\cos^2\theta\sin\theta-\sin^3\theta)$
Thus each component supplies its respective formula.
$\cos3\theta=\cos^3\theta-3\cos\theta\sin^2\theta,\quad\sin3\theta=3\cos^2\theta\sin\theta-\sin^3\theta$
For cosθ≠0 and cos3θ≠0, divide by cos³θ.
$\tan3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$

The sine/cosine identities hold for all θ; the tangent identity requires defined denominators.

Checks and common pitfalls: The tangent formula cannot be used when 1−3tan²θ=0.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Equate the expansion to cis(3θ).
$\cos3\theta+i\sin3\theta=(\cos^3\theta-3\cos\theta\sin^2\theta)+i(3\cos^2\theta\sin\theta-\sin^3\theta)$
Thus each component supplies its respective formula.
$\cos3\theta=\cos^3\theta-3\cos\theta\sin^2\theta,\quad\sin3\theta=3\cos^2\theta\sin\theta-\sin^3\theta$
For cosθ≠0 and cos3θ≠0, divide by cos³θ.
$\tan3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$

Think first. Reveal a hint when the class is ready.

2023 JM02