Multiple Angles by De Moivre

One number written two ways

By De Moivre’s theorem, $\left(\cos \theta + i \sin \theta\right)^{3} = \cos 3\theta + i \sin 3\theta$. The left side can also be multiplied out by the binomial theorem. Both results are the same complex number, so comparing them gives formulas for $\cos 3\theta$ and $\sin 3\theta$ in terms of $\cos \theta$ and $\sin \theta$.

Multiply out

Write c for $\cos \theta$ and s for $\sin \theta$. The binomial coefficients for a cube are 1, 3, 3 and 1, so $\left(c + \text{is}\right)^{3} = c^{3} + 3c^{2}\left(\text{is}\right) + 3c\left(\text{is}\right)^{2} + \left(\text{is}\right)^{3}$.

Now the powers of i. $\left(\text{is}\right)^{2} = i^{2}s^{2} = -s^{2}$, because $i^{2} = -1$. $\left(\text{is}\right)^{3} = i^{3}s^{3} = -i\left(s^{3}\right)$, because $i^{3} = -i$. So $\left(c + \text{is}\right)^{3} = c^{3} + i\left(3c^{2}s\right) - 3cs^{2} - i\left(s^{3}\right)$.

Gather the terms without i and the terms with i: $\left(c + \text{is}\right)^{3} = \left(c^{3} - 3cs^{2}\right) + i\left(3c^{2}s - s^{3}\right)$.

Match the parts

Two complex numbers are equal only when their real parts are equal and their imaginary parts are equal, because a real number can never make up for a multiple of i. So $\cos 3\theta + i \sin 3\theta = \left(c^{3} - 3cs^{2}\right) + i\left(3c^{2}s - s^{3}\right)$ gives two results: $\cos 3\theta = c^{3} - 3cs^{2}$ and $\sin 3\theta = 3c^{2}s - s^{3}$.

Check at $\theta = 30^\circ$, where $c = \frac{\sqrt{3}}{2}$ and $s = \frac{1}{2}$. Then $c^{3} - 3cs^{2} = \frac{3\sqrt{3}}{8} - \frac{3\sqrt{3}}{8} = 0$, which is cos 90°, and $3c^{2}s - s^{3} = \frac{9}{8} - \frac{1}{8} = 1$, which is sin 90°.

Only cosines

$\cos 3\theta = c^{3} - 3cs^{2}$ still has a sine in it. Replace $s^{2}$ by $1 - c^{2}$, from $\sin^{2}\theta + \cos^{2}\theta = 1$: $\cos 3\theta = c^{3} - 3c\left(1 - c^{2}\right) = c^{3} - 3c + 3c^{3} = 4c^{3} - 3c$.

So $\cos 3\theta = 4\cos^{3}\theta - 3\cos \theta$. Check at $\theta = 20^\circ$: $\cos 20^\circ \approx 0.939693$, so $4\cos^{3}20^\circ \approx 3.319078$ and $3\cos 20^\circ \approx 2.819078$, and the difference is 0.5, which is cos 60°.

The sine goes the same way with $c^{2}$ replaced by $1 - s^{2}$: $\sin 3\theta = 3\left(1 - s^{2}\right)s - s^{3} = 3s - 4s^{3}$. So $\sin 3\theta = 3\sin \theta - 4\sin^{3}\theta$.

The curve is $y = 4\cos^{3}\theta - 3\cos \theta$ for $\theta$ from 0° to 180°. The dots are $\cos 3\theta$ worked out separately every 30°: 1, 0, −1, 0, 1, 0, −1. Every dot lies on the curve.

The double angle the same way

At n = 2, $\left(c + \text{is}\right)^{2} = c^{2} + i\left(2cs\right) + i^{2}s^{2} = \left(c^{2} - s^{2}\right) + i\left(2cs\right)$. De Moivre says the same number is $\cos 2\theta + i \sin 2\theta$.

Matching the real parts gives $\cos 2\theta = \cos^{2}\theta - \sin^{2}\theta$, and matching the imaginary parts gives $\sin 2\theta = 2 \sin \theta \cos \theta$: the double-angle formulas.

The fourth power

The coefficients for a fourth power are 1, 4, 6, 4 and 1, and $i^{4} = 1$. So $\left(c + \text{is}\right)^{4} = c^{4} + i\left(4c^{3}s\right) - 6c^{2}s^{2} - i\left(4cs^{3}\right) + s^{4} = \left(c^{4} - 6c^{2}s^{2} + s^{4}\right) + i\left(4c^{3}s - 4cs^{3}\right)$.

Matching parts with $\cos 4\theta + i \sin 4\theta$: $\cos 4\theta = c^{4} - 6c^{2}s^{2} + s^{4}$ and $\sin 4\theta = 4c^{3}s - 4cs^{3}$. Replacing $s^{2}$ by $1 - c^{2}$ in the cosine: $\cos 4\theta = c^{4} - 6c^{2}\left(1 - c^{2}\right) + \left(1 - c^{2}\right)^{2} = c^{4} - 6c^{2} + 6c^{4} + 1 - 2c^{2} + c^{4} = 8c^{4} - 8c^{2} + 1$.

$\tan 3\theta$

Dividing the two cube results gives $\tan 3\theta = \frac{{3c^{2}s - s^{3}}}{{c^{3} - 3cs^{2}}}$. Divide the top and the bottom by $c^{3}$, and with $t = \tan \theta = \frac{s}{c}$ this is $\tan 3\theta = \frac{{3t - t^{3}}}{{1 - 3t^{2}}}$.

A cubic solved by an angle

Solve $8c^{3} - 6c - 1 = 0$. Halved, it is $4c^{3} - 3c = \frac{1}{2}$. If $c = \cos \theta$, the left side is $\cos 3\theta$, so the equation is $\cos 3\theta = \frac{1}{2}$.

$\cos 3\theta = \frac{1}{2}$ when $3\theta = 60^\circ$, 300° or 420°, so $\theta = 20^\circ$, 100° or 140°. The three roots of the cubic are $\cos 20^\circ \approx 0.940$, $\cos 100^\circ \approx -0.174$ and $\cos 140^\circ \approx -0.766$, and these three are different, so they are all the roots. Their sum is 0, matching the missing $c^{2}$ term.

The usual mistakes

Matching $\cos 3\theta$ with the i part. $\cos 3\theta$ stands alone, with no i, so it matches the real part $c^{3} - 3cs^{2}$; the i part matches $\sin 3\theta$.

Swapping the coefficients: $3\cos^{3}\theta - 4\cos \theta$. At $\theta = 0$ that gives 3 − 4 = −1, but cos 0° = 1. The 4 belongs to the cube.

Dropping the $3cs^{2}$ term, which leaves $\cos^{3}\theta - 3\cos \theta$. Replacing $s^{2}$ by $1 - c^{2}$ adds $3c^{3}$ to the $c^{3}$ already there.

Forgetting that $i^{2} = -1$ and $i^{3} = -i$. Without them the signs of the $3cs^{2}$ and $s^{3}$ terms come out wrong.

Taking $c^{2} - s^{2}$ as $\sin 2\theta$. It carries no i, so it matches $\cos 2\theta$; the i part, 2cs, is $\sin 2\theta$.

A cam with a cubic lift

In the application below, the lift is $h = 8\cos^{3}\theta - 6\cos \theta + 2$, which is $2\left(4\cos^{3}\theta - 3\cos \theta\right) + 2 = 2\cos 3\theta + 2$. A lift of 3 cm needs $\cos 3\theta = \frac{1}{2}$, the equation just solved, and the smallest positive shaft angle is 20°, written $\frac{\pi}{9}$ in radians.