4.02q - De Moivre's theorem for multiple angles

4.02q - De Moivre's theorem for multiple angles

Use de Moivre's theorem to turn a complex power into a multiple-angle identity. You will derive formulae for sine, cosine and tangent, explain the conditions on a quotient, and use a trigonometric identity to solve a polynomial equation exactly.

De Moivre's theorem

How can the cubic equation t3−3t2−3t+1=0t^3 - 3t^2 - 3t + 1 = 0 have roots involving familiar angles, even though it contains no trigonometry? The bridge is a complex number on the unit circle.

Take z=cos⁡θ+isin⁡θ=eiθz = \cos\theta + i\sin\theta = e^{i\theta}. Multiplying three copies adds three arguments, so

z3=e3iθ=cos⁡3θ+isin⁡3θz^3 = e^{3i\theta} = \cos 3\theta + i\sin 3\theta

But expanding (cos⁡θ+isin⁡θ)3(\cos\theta + i\sin\theta)^3 gives powers of cos⁡θ\cos\theta and sin⁡θ\sin\theta. The same complex number therefore has two descriptions. Equating their real and imaginary parts gives two real identities at once: complex numbers do double duty.

You already know

Euler's formula is eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos\theta + i\sin\theta. Multiplying complex numbers multiplies their moduli and adds their arguments. Also i2=−1i^2 = -1, i3=−ii^3 = -i and i4=1i^4 = 1.

The same reasoning works for any positive integer power and any modulus. If z=reiθz = re^{i\theta}, then zn=rneinθz^n = r^n e^{in\theta}.

De Moivre's theorem

For real θ\theta, r>0r > 0 and integer nn,

{r(cos⁡θ+isin⁡θ)}n=rn(cos⁡nθ+isin⁡nθ)\{r(\cos\theta + i\sin\theta)\}^n = r^n(\cos n\theta + i\sin n\theta)

For positive integer nn, this also holds when r=0r = 0. For n=0n = 0 or a negative integer, the base must be non-zero.

In the formula booklet.

For n=0n = 0, both sides are 1. For n=−mn = -m, take the reciprocal of the positive-power result: the modulus becomes r−mr^{-m} and the angle becomes −mθ-m\theta. Do not apply this as a single-valued formula for fractional powers: finding all roots needs all the arguments of the original number.

A useful surprise is already visible: (cos⁡θ+isin⁡θ)n(\cos\theta + i\sin\theta)^n, no matter how long its binomial expansion looks, always has modulus 1. Every real and imaginary term in that expansion must combine to land back on the unit circle.

Key idea

Expand a complex power in two ways: de Moivre gives the multiple angle; the binomial theorem gives powers of the original sine and cosine.

With a negative exponent, check the modulus and the angle separately: both must follow the power law.

Sine and cosine of multiple angles

To derive an identity, expand (c+is)n(c + is)^n, where c=cos⁡θc = \cos\theta and s=sin⁡θs = \sin\theta, then compare it with cos⁡nθ+isin⁡nθ\cos n\theta + i\sin n\theta.

The real terms contain even powers of ii; the imaginary terms contain odd powers. If the question wants only powers of cosine, replace s2s^2 by 1−c21 - c^2. If it wants only powers of sine, replace c2c^2 by 1−s21 - s^2. These replacements do not involve division, so the resulting sine and cosine identities hold for every real θ\theta.

Worked example 1: triple-angle identities

Question. Use de Moivre's theorem to show that

cos⁡3θ=4cos⁡3θ−3cos⁡θandsin⁡3θ=3sin⁡θ−4sin⁡3θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta \quad\text{and}\quad \sin 3\theta = 3\sin\theta - 4\sin^3\theta

Plan. Expand the cube, equate parts, then use c2+s2=1c^2 + s^2 = 1 to leave one trigonometric ratio in each identity.

cos⁡3θ+isin⁡3θ=(c+is)3de Moivre=c3+3ic2s+3i2cs2+i3s3binomial expansion=(c3−3cs2)+i(3c2s−s3)i2=−1, i3=−i\begin{aligned} \cos 3\theta + i\sin 3\theta &= (c + is)^3 && \text{de Moivre} \\ &= c^3 + 3ic^2s + 3i^2cs^2 + i^3s^3 && \text{binomial expansion} \\ &= (c^3 - 3cs^2) + i(3c^2s - s^3) && i^2=-1,\ i^3=-i \end{aligned}

Why this step? Collecting the terms without ii separately from the coefficients of ii lets you equate two real quantities on each side.

cos⁡3θ=c3−3c(1−c2)=4c3−3creal partssin⁡3θ=3(1−s2)s−s3=3s−4s3imaginary parts\begin{aligned} \cos 3\theta &= c^3 - 3c(1-c^2) = 4c^3 - 3c && \text{real parts} \\ \sin 3\theta &= 3(1-s^2)s - s^3 = 3s - 4s^3 && \text{imaginary parts} \end{aligned}

Answer. Both required identities follow, for all real θ\theta.

Check. At θ=π6\theta = \frac{\pi}{6}, the sine identity gives 3(12)−4(12)3=1=sin⁡π23(\frac12) - 4(\frac12)^3 = 1 = \sin\frac{\pi}{2}; the cosine identity gives 4(32)3−3(32)=04(\frac{\sqrt3}{2})^3 - 3(\frac{\sqrt3}{2}) = 0.

Worked example 2 changes only the multiplier, from 3 to 5. The method stays the same; the extra binomial terms make careful collection more important.

Worked example 2: quintuple-angle identities

Question. Use de Moivre's theorem to express cos⁡5θ\cos 5\theta as a polynomial in cos⁡θ\cos\theta, and sin⁡5θ\sin 5\theta as a polynomial in sin⁡θ\sin\theta.

Plan. Expand the fifth power, collect parts, then eliminate the squared complementary ratio.

cos⁡5θ+isin⁡5θ=(c+is)5de Moivre=c5+5ic4s−10c3s2−10ic2s3+5cs4+is5expand; reduce powers of i\begin{aligned} \cos5\theta+i\sin5\theta &= (c+is)^5 && \text{de Moivre} \\ &= c^5+5ic^4s-10c^3s^2 \\ &\quad -10ic^2s^3+5cs^4+is^5 && \text{expand; reduce powers of }i \end{aligned}

Thus

cos⁡5θ=c5−10c3s2+5cs4real parts=c5−10c3(1−c2)+5c(1−c2)2s2=1−c2=16c5−20c3+5ccollect powerssin⁡5θ=5c4s−10c2s3+s5imaginary parts=5(1−s2)2s−10(1−s2)s3+s5c2=1−s2=16s5−20s3+5scollect powers\begin{aligned} \cos5\theta &= c^5-10c^3s^2+5cs^4 && \text{real parts} \\ &= c^5-10c^3(1-c^2)+5c(1-c^2)^2 && s^2=1-c^2 \\ &=16c^5-20c^3+5c && \text{collect powers} \\ \sin5\theta &=5c^4s-10c^2s^3+s^5 && \text{imaginary parts} \\ &=5(1-s^2)^2s-10(1-s^2)s^3+s^5 && c^2=1-s^2 \\ &=16s^5-20s^3+5s && \text{collect powers} \end{aligned}

Why this step? Replacing s2s^2 in the real part and c2c^2 in the imaginary part leaves exactly the single ratio requested in each answer.

Answer. cos⁡5θ=16cos⁡5θ−20cos⁡3θ+5cos⁡θ\cos5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta, and sin⁡5θ=16sin⁡5θ−20sin⁡3θ+5sin⁡θ\sin5\theta=16\sin^5\theta-20\sin^3\theta+5\sin\theta.

Check. At θ=π6\theta = \frac{\pi}{6}, the sine polynomial is 12−52+52=12=sin⁡5π6\frac12-\frac52+\frac52=\frac12=\sin\frac{5\pi}{6}. The cosine polynomial is −32=cos⁡5π6-\frac{\sqrt3}{2}=\cos\frac{5\pi}{6}.

The derived triple- and quintuple-angle identities are not printed in the formula booklet: derive them from de Moivre's theorem when needed.

Watch out

Wrong: the imaginary part of c3+3ic2s−3cs2−is3c^3 + 3ic^2s - 3cs^2 - is^3 is 3ic2s−is33ic^2s - is^3. The imaginary part is the real coefficient of ii. Right: equate sin⁡3θ=3c2s−s3\sin3\theta = 3c^2s - s^3, with no ii left.

For an even multiplier, both s2s^2 and s4s^4 can enter the real part. Replace each using s2=1−c2s^2=1-c^2 before collecting the cosine powers.

Tangent identities and polynomial roots

Divide the imaginary part by the real part to obtain tangent, but check that the real part is non-zero before dividing.

For example, tan⁡3θ=sin⁡3θ/cos⁡3θ\tan3\theta=\sin3\theta/\cos3\theta requires cos⁡3θ≠0\cos3\theta\ne0. To rewrite the result using tan⁡θ\tan\theta, you also divide by a power of cos⁡θ\cos\theta, requiring cos⁡θ≠0\cos\theta\ne0. These are conditions on the identity, not optional fine print.

Triple-angle tangent

Where cos⁡θ≠0\cos\theta\ne0 and cos⁡3θ≠0\cos3\theta\ne0,

tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}

Not in the formula booklet: learn it.

Here is its derivation from the expansion in the previous part:

tan⁡3θ=3c2s−s3c3−3cs2imaginary part over real part=3c2sc3−s3c3c3c3−3cs2c3divide each term by c3=3tan⁡θ−tan⁡3θ1−3tan⁡2θs/c=tan⁡θ\begin{aligned} \tan3\theta&=\frac{3c^2s-s^3}{c^3-3cs^2} && \text{imaginary part over real part} \\ &=\frac{\dfrac{3c^2s}{c^3}-\dfrac{s^3}{c^3}}{\dfrac{c^3}{c^3}-\dfrac{3cs^2}{c^3}} && \text{divide each term by }c^3 \\ &=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta} && s/c=\tan\theta \end{aligned}

Why this step? Dividing every term by the same power of cc turns every remaining s/cs/c into tangent. Showing that line makes the derivation complete.

Now turn a trigonometric equation into a polynomial. Choosing 0≤θ<π0\le\theta<\pi, apart from π2\frac\pi2, makes t=tan⁡θt=\tan\theta take every real value exactly once.

Worked example 3: a quadratic from a double angle

Question. By considering tan⁡2θ=1\tan2\theta=1, determine the exact roots of t2+2t−1=0t^2+2t-1=0 in the form t=tan⁡αt=\tan\alpha, where 0≤α<π0\le\alpha<\pi.

Plan. Substitute t=tan⁡θt=\tan\theta into the double-angle tangent formula and list the angles in the given interval.

tan⁡2θ=2t1−t2=1double-angle identity2t=1−t2non-zero denominatort2+2t−1=0rearrange2θ=π4+kπtan⁡2θ=1θ=π8, 5π80≤θ<π\begin{aligned} \tan2\theta&=\frac{2t}{1-t^2}=1 && \text{double-angle identity} \\ 2t&=1-t^2 && \text{non-zero denominator} \\ t^2+2t-1&=0 && \text{rearrange} \\ 2\theta&=\frac\pi4+k\pi && \tan2\theta=1 \\ \theta&=\frac\pi8,\ \frac{5\pi}8 && 0\le\theta<\pi \end{aligned}

Why this step? Solving for the angle first gives every tangent value without guessing a factor of the polynomial.

Answer. t=tan⁡π8t=\tan\frac\pi8 or t=tan⁡5π8t=\tan\frac{5\pi}8.

Check. The quadratic formula gives −1±2-1\pm\sqrt2, approximately 0.4140.414 and −2.414-2.414, matching those tangents. Neither has 1−t2=01-t^2=0, so neither was introduced by multiplying through.

Worked example 4 changes only the multiplier from 2 to 3. There are now three angle solutions and a cubic in tt.

Worked example 4: the opening cubic

Question. By considering tan⁡3θ=1\tan3\theta=1, determine the exact roots of t3−3t2−3t+1=0t^3-3t^2-3t+1=0 in the form t=tan⁡αt=\tan\alpha, where 0≤α<π0\le\alpha<\pi.

Plan. Apply the triple-angle identity, rearrange, then solve the angle equation over one interval of length π\pi.

3t−t31−3t2=1t=tan⁡θ3t−t3=1−3t2multiply throught3−3t2−3t+1=0rearrange3θ=π4+kπtan⁡3θ=1θ=π12, 5π12, 3π40≤θ<π\begin{aligned} \frac{3t-t^3}{1-3t^2}&=1 && t=\tan\theta \\ 3t-t^3&=1-3t^2 && \text{multiply through} \\ t^3-3t^2-3t+1&=0 && \text{rearrange} \\ 3\theta&=\frac\pi4+k\pi && \tan3\theta=1 \\ \theta&=\frac\pi{12},\ \frac{5\pi}{12},\ \frac{3\pi}4 && 0\le\theta<\pi \end{aligned}

Why this step? Increasing θ\theta by π/3\pi/3 leaves tan⁡3θ\tan3\theta unchanged, but produces a different value of tan⁡θ\tan\theta. That supplies three distinct roots.

Answer. t=tan⁡π12t=\tan\frac\pi{12}, tan⁡5π12\tan\frac{5\pi}{12} or tan⁡3π4\tan\frac{3\pi}4.

Check. These are 2−32-\sqrt3, 2+32+\sqrt3 and −1-1. The factorisation (t+1)(t2−4t+1)(t+1)(t^2-4t+1) gives exactly the same roots. If 1−3t2=01-3t^2=0, the numerator is 83t≠0\frac83t\ne0, so no excluded denominator value is a root of the cubic.

The double-angle tangent formula used above is in the formula booklet as the compound-angle formula with the two angles equal; the specialised formula is not separately printed. The polynomial relationships here are derived consequences, also not printed.

Watch out

Wrong: tan⁡3θ=3tan⁡θ\tan3\theta=3\tan\theta. De Moivre multiplies the angle inside sine and cosine; it does not multiply their values. Right: expand first, equate parts, then take their quotient.

The expanded quotient is also the part of the argument where a concise answer can become too concise: the conversion to tangent must remain visible.

Exam tip

In a detailed-reasoning derivation, show each term in both the numerator and denominator divided by the required power of cos⁡θ\cos\theta. Merely writing “divide by cos⁡nθ\cos^n\theta” can leave the final accuracy mark unearned.

The same chain works with multiplier 4; the angle equation then makes four distinct tangent values appear over an interval of length π\pi.

Quick Recap

  • De Moivre turns an integer power into a multiple angle, while raising the modulus to the same power.
  • Expand the same power binomially and equate real and imaginary parts to derive cosine and sine identities.
  • Use sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 to express the result using one ratio.
  • Tangent comes from imaginary part divided by real part; show the division by a power of cosine and keep its conditions.
  • An angle equation can reveal all the real roots of a polynomial in tan⁡θ\tan\theta.

The opening cubic has roots tan⁡π12\tan\frac\pi{12}, tan⁡5π12\tan\frac{5\pi}{12} and tan⁡3π4\tan\frac{3\pi}4: its hidden equation is tan⁡3θ=1\tan3\theta=1.