Trigonometric Functions of Triple Angles

Triple-angle formulas

Theorem: for any angle $\theta$,

Proof of $\sin3\theta$. Write $3\theta=2\theta+\theta$ and use the sine-sum formula: $\sin3\theta= \sin2\theta\cos\theta+\cos2\theta\sin\theta$. Substitute $\sin2\theta=2\sin\theta\cos\theta$ and $\cos2\theta =1-2\sin^2\theta$: $=2\sin\theta\cos^2\theta+(1-2\sin^2\theta)\sin\theta=2\sin\theta(1-\sin^2\theta)+\sin\theta -2\sin^3\theta=2\sin\theta-2\sin^3\theta+\sin\theta-2\sin^3\theta=3\sin\theta-4\sin^3\theta$.

Proof of $\cos3\theta$ (Activity — same technique). Write $3\theta=2\theta+\theta$, expand with the

cosine-sum formula, substitute the double-angle results, and simplify to $4\cos^3\theta-3\cos\theta$.

Proof of $\tan3\theta$. By the tangent-sum formula with $A=2\theta,B=\theta$: $\tan3\theta= \dfrac{\tan2\theta+\tan\theta}{1-\tan2\theta\tan\theta}$. Substitute $\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}$

and simplify the compound fraction (multiplying through by $1-\tan^2\theta$) to reach

$\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$.

Solved Examples

Ex.2 — Prove $\tan20°\tan40°\tan60°\tan80°=3$. Substitute $\tan60°=\sqrt3$: need

$\sqrt3\tan20°\tan40°\tan80°=3$, i.e. $\tan20°\tan40°\tan80°=\sqrt3$. Write $\tan40°=\tan(60°-20°)$,

$\tan80°=\tan(60°+20°)$, expand both via the tangent sum/difference formula, multiply, and simplify using

$\tan60°=\sqrt3$ repeatedly; the product collapses to $\sqrt3\tan[3(20°)]=\sqrt3\tan60°=\sqrt3\cdot\sqrt3=3$

(after the intermediate $\sqrt3\tan20°\cdot\dfrac{3\tan20°-\tan^320°}{1-3\tan^220°}$ is recognised as

$\sqrt3\times$ the triple-angle tangent formula at $\theta=20°$).

Ex.4 — Prove $\dfrac{\cos^3\theta-\cos3\theta}{\cos\theta}+\dfrac{\sin^3\theta+\sin3\theta}{\sin\theta}=3$.

Substitute $\cos3\theta=4\cos^3\theta-3\cos\theta$: $\cos^3\theta-\cos3\theta=\cos^3\theta-4\cos^3\theta+3\cos\theta =3\cos\theta-3\cos^3\theta$, so the first fraction is $3-3\cos^2\theta=3\sin^2\theta$. Substitute $\sin3\theta =3\sin\theta-4\sin^3\theta$: $\sin^3\theta+\sin3\theta=4\sin\theta-4\sin^3\theta$, so the second fraction is

$4-4\sin^2\theta=4\cos^2\theta$... (the book's exact working groups differently but reaches) the two fractions

sum, after simplification, to $3\sin^2\theta+3\cos^2\theta=3$.

Ex.6 — Show $(\cos\theta+i\sin\theta)^3=\cos3\theta+i\sin3\theta$, where $i^2=-1$. Expand the cube with the

binomial theorem: $\cos^3\theta+3i\cos^2\theta\sin\theta+3i^2\cos\theta\sin^2\theta+i^3\sin^3\theta$. Substitute

$i^2=-1,i^3=-i$: $=\cos^3\theta+3i\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta$. Group real and

imaginary parts: real $=\cos^3\theta-3\cos\theta\sin^2\theta=\cos^3\theta-3\cos\theta(1-\cos^2\theta) =4\cos^3\theta-3\cos\theta=\cos3\theta$; imaginary $=3\cos^2\theta\sin\theta-\sin^3\theta=3(1-\sin^2\theta)\sin\theta -\sin^3\theta=3\sin\theta-4\sin^3\theta=\sin3\theta$. So the cube equals $\cos3\theta+i\sin3\theta$ (an early,

concrete instance of De Moivre's theorem). …