Multiple Angles by De Moivre

One number written two ways

By De Moivre’s theorem, ${\left(\cos \theta +i\sin \theta \right)}^3=\cos 3\theta +i\sin 3\theta$(cos θ + i sin θ)³ = cos 3θ + i sin 3θ. 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 $\cos 3\theta$cos 3θ and $\sin 3\theta$sin 3θ in terms of $\cos \theta$cos θ and $\sin \theta$sin θ.

Multiply out

Write c for $\cos \theta$cos θ and s for $\sin \theta$sin θ. The binomial coefficients for a cube are 1, 3, 3 and 1, so ${\left(c+\text{is}\right)}^3=c^3+3c^2\left(\text{is}\right)+3c{\left(\text{is}\right)}^2+{\left(\text{is}\right)}^3$(c + is)³ = c³ + 3c²(is) + 3c(is)² + (is)³.

Now the powers of i. ${\left(\text{is}\right)}^2=i^2{s}^2=-s^2$(is)² = i²s² = −s², because $i^2=-1$i² = −1. ${\left(\text{is}\right)}^3=i^3{s}^3=-i\left(s^3\right)$(is)³ = i³s³ = −i(s³), because $i^3=-i$i³ = −i. So ${\left(c+\text{is}\right)}^3=c^3+i\left(3c^{2}s\right)-3c{s}^2-i\left(s^3\right)$(c + is)³ = c³ + i(3c²s) − 3cs² − i(s³).

Gather the terms without i and the terms with i: ${\left(c+\text{is}\right)}^3=\left(c^3-3c{s}^2\right)+i\left(3c^{2}s-s^3\right)$(c + is)³ = (c³ − 3cs²) + i(3c²s − s³).

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 $\cos 3\theta +i\sin 3\theta =\left(c^3-3c{s}^2\right)+i\left(3c^{2}s-s^3\right)$cos 3θ + i sin 3θ = (c³ − 3cs²) + i(3c²s − s³) gives two results: $\cos 3\theta =c^3-3c{s}^2$cos 3θ = c³ − 3cs² and $\sin 3\theta =3c^{2}s-s^3$sin 3θ = 3c²s − s³.

Check at $\theta =30^{\circ }$θ = 30°, where $c=\frac{\sqrt{3}}{2}$c = √3/2 and $s=\frac{1}{2}$s = 1/2. Then $c^3-3c{s}^2=\frac{3\sqrt{3}}{8}-\frac{3\sqrt{3}}{8}=0$c³ − 3cs² = 3√3/8 − 3√3/8 = 0, which is cos 90°, and $3c^{2}s-s^3=\frac{9}{8}-\frac{1}{8}=1$3c²s − s³ = 9/8 − 1/8 = 1, which is sin 90°.

Only cosines

$\cos 3\theta =c^3-3c{s}^2$cos 3θ = c³ − 3cs² still has a sine in it. Replace $s^2$s² by $1-c^2$1 − c², from ${\sin }^2\theta +{\cos }^2\theta =1$sin²θ + cos²θ = 1: $\cos 3\theta =c^3-3c\left(1-c^2\right)=c^3-3c+3c^3=4c^3-3c$cos 3θ = c³ − 3c(1 − c²) = c³ − 3c + 3c³ = 4c³ − 3c.

So $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$cos 3θ = 4cos³θ − 3cos θ. Check at $\theta =20^{\circ }$θ = 20°: $\cos 20^{\circ }\approx 0.939693$cos 20° ≈ 0.939693, so $4{\cos }^{3}20^{\circ }\approx 3.319078$4cos³20° ≈ 3.319078 and $3\cos 20^{\circ }\approx 2.819078$3cos 20° ≈ 2.819078, and the difference is 0.5, which is cos 60°.

The sine goes the same way with $c^2$c² replaced by $1-s^2$1 − s²: $\sin 3\theta =3\left(1-s^2\right)s-s^3=3s-4s^3$sin 3θ = 3(1 − s²)s − s³ = 3s − 4s³. So $\sin 3\theta =3\sin \theta -4{\sin }^3\theta$sin 3θ = 3sin θ − 4sin³θ.

The curve is $y=4{\cos }^3\theta -3\cos \theta$y = 4cos³θ − 3cos θ for $\theta$θ from 0° to 180°. The dots are $\cos 3\theta$cos 3θ 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, ${\left(c+\text{is}\right)}^2=c^2+i\left(2cs\right)+i^2{s}^2=\left(c^2-s^2\right)+i\left(2cs\right)$(c + is)² = c² + i(2cs) + i²s² = (c² − s²) + i(2cs). De Moivre says the same number is $\cos 2\theta +i\sin 2\theta$cos 2θ + i sin 2θ.

Matching the real parts gives $\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$cos 2θ = cos²θ − sin²θ, and matching the imaginary parts gives $\sin 2\theta =2\sin \theta \cos \theta$sin 2θ = 2 sin θ cos θ: the double-angle formulas.

The fourth power

The coefficients for a fourth power are 1, 4, 6, 4 and 1, and $i^4=1$i⁴ = 1. So ${\left(c+\text{is}\right)}^4=c^4+i\left(4c^{3}s\right)-6c^2{s}^2-i\left(4c{s}^3\right)+s^4=\left(c^4-6c^2{s}^2+s^4\right)+i\left(4c^{3}s-4c{s}^3\right)$(c + is)⁴ = c⁴ + i(4c³s) − 6c²s² − i(4cs³) + s⁴ = (c⁴ − 6c²s² + s⁴) + i(4c³s − 4cs³).

Matching parts with $\cos 4\theta +i\sin 4\theta$cos 4θ + i sin 4θ: $\cos 4\theta =c^4-6c^2{s}^2+s^4$cos 4θ = c⁴ − 6c²s² + s⁴ and $\sin 4\theta =4c^{3}s-4c{s}^3$sin 4θ = 4c³s − 4cs³. Replacing $s^2$s² by $1-c^2$1 − c² in the cosine: $\cos 4\theta =c^4-6c^2\left(1-c^2\right)+{\left(1-c^2\right)}^2=c^4-6c^2+6c^4+1-2c^2+c^4=8c^4-8c^2+1$cos 4θ = c⁴ − 6c²(1 − c²) + (1 − c²)² = c⁴ − 6c² + 6c⁴ + 1 − 2c² + c⁴ = 8c⁴ − 8c² + 1.

$\tan 3\theta$tan 3θ

Dividing the two cube results gives $\tan 3\theta =\frac{3c^{2}s-s^3}{c^3-3c{s}^2}$tan 3θ = (3c²s − s³)/(c³ − 3cs²). Divide the top and the bottom by $c^3$c³, and with $t=\tan \theta =\frac{s}{c}$t = tan θ = s/c this is $\tan 3\theta =\frac{3t-t^3}{1-3t^2}$tan 3θ = (3t − t³)/(1 − 3t²).

A cubic solved by an angle

Solve $8c^3-6c-1=0$8c³ − 6c − 1 = 0. Halved, it is $4c^3-3c=\frac{1}{2}$4c³ − 3c = 1/2. If $c=\cos \theta$c = cos θ, the left side is $\cos 3\theta$cos 3θ, so the equation is $\cos 3\theta =\frac{1}{2}$cos 3θ = 1/2.

$\cos 3\theta =\frac{1}{2}$cos 3θ = 1/2 when $3\theta =60^{\circ }$3θ = 60°, 300° or 420°, so $\theta =20^{\circ }$θ = 20°, 100° or 140°. The three roots of the cubic are $\cos 20^{\circ }\approx 0.940$cos 20° ≈ 0.940, $\cos 100^{\circ }\approx -0.174$cos 100° ≈ −0.174 and $\cos 140^{\circ }\approx -0.766$cos 140° ≈ −0.766, and these three are different, so they are all the roots. Their sum is 0, matching the missing $c^2$c² term.

The usual mistakes

Matching $\cos 3\theta$cos 3θ with the i part. $\cos 3\theta$cos 3θ stands alone, with no i, so it matches the real part $c^3-3c{s}^2$c³ − 3cs²; the i part matches $\sin 3\theta$sin 3θ.

Swapping the coefficients: $3{\cos }^3\theta -4\cos \theta$3cos³θ − 4cos θ. At $\theta =0$θ = 0 that gives 3 − 4 = −1, but cos 0° = 1. The 4 belongs to the cube.

Dropping the $3c{s}^2$3cs² term, which leaves ${\cos }^3\theta -3\cos \theta$cos³θ − 3cos θ. Replacing $s^2$s² by $1-c^2$1 − c² adds $3c^3$3c³ to the $c^3$c³ already there.

Forgetting that $i^2=-1$i² = −1 and $i^3=-i$i³ = −i. Without them the signs of the $3c{s}^2$3cs² and $s^3$s³ terms come out wrong.

Taking $c^2-s^2$c² − s² as $\sin 2\theta$sin 2θ. It carries no i, so it matches $\cos 2\theta$cos 2θ; the i part, 2cs, is $\sin 2\theta$sin 2θ.

A cam with a cubic lift

In the application below, the lift is $h=8{\cos }^3\theta -6\cos \theta +2$h = 8cos³θ − 6cos θ + 2, which is $2\left(4{\cos }^3\theta -3\cos \theta \right)+2=2\cos 3\theta +2$2(4cos³θ − 3cos θ) + 2 = 2cos 3θ + 2. A lift of 3 cm needs $\cos 3\theta =\frac{1}{2}$cos 3θ = 1/2, the equation just solved, and the smallest positive shaft angle is 20°, written $\frac{\pi }{9}$π/9 in radians.