Compound Angles and 3D Trigonometry

≈ 25 min

Compound-angle identities

\cos(A-B) = \cos A\cos B + \sin A\sin B \qquad \cos(A+B) = \cos A\cos B - \sin A\sin B

\sin(A+B) = \sin A\cos B + \cos A\sin B \qquad \sin(A-B) = \sin A\cos B - \cos A\sin B

Putting $B = A$ gives the double-angle formulae:

\sin 2A = 2\sin A\cos A, \qquad \cos 2A = \cos^2 A - \sin^2 A = 1 - 2\sin^2 A = 2\cos^2 A - 1

Worked example: an exact value

\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ\cos 30^\circ + \sin 45^\circ\sin 30^\circ = \frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} + \frac{\sqrt2}{2}\cdot\frac12 = \frac{\sqrt6 + \sqrt2}{4}

Numerically that is $\approx 0{,}966$ , which matches $\cos 15^\circ$ on a calculator.

Three-dimensional problems

In a 3D sketch, first identify the right-angled triangle in a vertical plane and the triangle in the horizontal plane, then link them by a shared side.

Worked example: a cellphone mast

A mast $PQ$ stands vertically at $P$ . From a point $A$ on level ground, $PA = 40$ m and the angle of elevation of the top $Q$ is $32^\circ$ . In right-angled $\triangle APQ$ :

\tan 32^\circ = \frac{PQ}{40} \implies PQ = 40\tan 32^\circ \approx 40(0{,}6249) \approx 25{,}0\ \text{m}

When the horizontal triangle is not right-angled, use the tools you already know:

\frac{a}{\sin A} = \frac{b}{\sin B}, \qquad a^2 = b^2 + c^2 - 2bc\cos A, \qquad \text{Area} = \tfrac12 bc\sin A

Simplify $\sin(90^\circ + x)$ using a compound-angle identity.

$\cos x$

$-\cos x$

$\sin x$

$-\sin x$

A solar panel is tilted at an angle $\theta$ where $\sin\theta = \dfrac{3}{5}$ and $\theta$ is acute. Calculate the value of $\sin 2\theta$ as a decimal.

Two Eskom pylons $B$ and $C$ stand on level ground and are viewed from a substation at $A$ . The horizontal distances are $AB = 12$ m and $AC = 15$ m, with $B\hat{A}C = 60^\circ$ . Calculate the distance $BC$ in metres, correct to two decimal places.