CP 2.8 - Multiple-angle formulae

CP 2.8 - Multiple-angle formulae

Use de Moivre's theorem to turn the sine, cosine or tangent of a multiple angle into powers of trigonometric functions of the original angle. You will choose real or imaginary parts, simplify the resulting polynomial and keep the restrictions that matter when dividing.

Real parts give multiple cosines

Why does x=cos⁡π9x = \cos\frac{\pi}{9} satisfy the cubic equation 8x3−6x−1=08x^3 - 6x - 1 = 0? A link between a triple angle and a cube will explain it. The link is an identity, not an instruction to confuse the two.

You already know

De Moivre gives (cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ(\cos\theta + \mathrm{i}\sin\theta)^n = \cos n\theta + \mathrm{i}\sin n\theta for integer nn. From A-level Mathematics, use the binomial expansion and sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.

The notation carries a big difference. cos⁡3θ\cos^3\theta means (cos⁡θ)3(\cos\theta)^3; cos⁡3θ\cos 3\theta means find the cosine after tripling the angle. Over one complete turn, cos⁡3θ\cos 3\theta completes three oscillations while cos⁡3θ\cos^3\theta completes one. A cubic polynomial can nevertheless connect them exactly.

[DIAGRAM: asset_name: Multiple angle compared with a power; asset_slug: 9fm0-cp09-angle-or-power; caption: Cosine of a triple angle and the cube of a cosine are different functions; description: On axes from theta equals zero to two pi and vertical values minus one to one, a blue curve cos three theta completes three oscillations and a clay curve cos cubed theta completes one. Labels identify three oscillations and cubing the height.]
Cosine of a triple angle and the cube of a cosine are different functions

Write c=cos⁡θc = \cos\theta and s=sin⁡θs = \sin\theta temporarily. Expand (c+is)n(c + \mathrm{i}s)^n, replace the powers of i\mathrm{i}, then collect its real terms. De Moivre identifies those terms as cos⁡nθ\cos n\theta.

Reading the real part

For real θ\theta and a positive integer nn,

cos⁡nθ=Re⁡((cos⁡θ+isin⁡θ)n)\cos n\theta = \operatorname{Re}\bigl((\cos\theta + \mathrm{i}\sin\theta)^n\bigr)

Even powers of i\mathrm{i} contribute to the real part. The resulting multiple-angle identities are derived from de Moivre; they are not separately printed in the formula booklet.

Not in the formula booklet: learn it.

The first example extracts the real part of a cubic expansion; its final form uses cos⁡θ\cos\theta alone.

Worked example 1: the triple-angle cosine

Question. Use de Moivre's theorem to show that cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta.

Plan. Expand the cube, equate real parts and replace sin⁡2θ\sin^2\theta by 1−cos⁡2θ1 - \cos^2\theta.

With c=cos⁡θc = \cos\theta and s=sin⁡θs = \sin\theta,

cos⁡3θ+isin⁡3θ=(c+is)3de Moivre=c3+3c2is+3ci2s2+i3s3binomial expansion=c3−3cs2+i(3c2s−s3)i2=−1, i3=−i\begin{aligned} \cos 3\theta + \mathrm{i}\sin 3\theta &= (c + \mathrm{i}s)^3 && \text{de Moivre} \\ &= c^3 + 3c^2\mathrm{i}s + 3c\mathrm{i}^2s^2 + \mathrm{i}^3s^3 && \text{binomial expansion} \\ &= c^3 - 3cs^2 + \mathrm{i}(3c^2s - s^3) && \mathrm{i}^2=-1,\ \mathrm{i}^3=-\mathrm{i} \end{aligned}

Equating real parts,

cos⁡3θ=c3−3cs2real terms=c3−3c(1−c2)s2=1−c2=4c3−3ccollect powers\begin{aligned} \cos 3\theta &= c^3 - 3cs^2 && \text{real terms} \\ &= c^3 - 3c(1-c^2) && s^2=1-c^2 \\ &= 4c^3 - 3c && \text{collect powers} \end{aligned}

Why this step? Replacing s2s^2 removes the other trigonometric function, giving the requested polynomial in cos⁡θ\cos\theta.

Answer. cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta, as required.

Check. At θ=π3\theta = \frac{\pi}{3}, the right side is 4(12)3−3(12)=−1=cos⁡π4(\frac12)^3 - 3(\frac12) = -1 = \cos\pi.

Now change only the integer multiplier, from 33 to 55. The expansion is longer, but the real terms still come from even powers of i\mathrm{i}.

Worked example 2: the quintuple-angle cosine

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

Plan. Take the real terms of (c+is)5(c + \mathrm{i}s)^5, then remove s2s^2 and s4s^4 using s2=1−c2s^2 = 1-c^2.

cos⁡5θ=Re⁡((c+is)5)de Moivre=c5+10c3i2s2+5ci4s4even powers only=c5−10c3s2+5cs4i2=−1, i4=1=c5−10c3(1−c2)+5c(1−c2)2eliminate s=c5−10c3+10c5+5c−10c3+5c5expand=16c5−20c3+5ccollect powers\begin{aligned} \cos 5\theta &= \operatorname{Re}\bigl((c+\mathrm{i}s)^5\bigr) && \text{de Moivre} \\ &= c^5 + 10c^3\mathrm{i}^2s^2 + 5c\mathrm{i}^4s^4 && \text{even powers only} \\ &= c^5 - 10c^3s^2 + 5cs^4 && \mathrm{i}^2=-1,\ \mathrm{i}^4=1 \\ &= c^5 - 10c^3(1-c^2) + 5c(1-c^2)^2 && \text{eliminate }s \\ &= c^5 - 10c^3 + 10c^5 + 5c - 10c^3 + 5c^5 && \text{expand} \\ &= 16c^5 - 20c^3 + 5c && \text{collect powers} \end{aligned}

Why this step? The coefficient 1010 belongs to the term containing s2s^2, while the coefficient 55 belongs to the term containing s4s^4. Writing those terms before simplifying prevents a coefficient swap.

Answer. cos⁡5θ=16cos⁡5θ−20cos⁡3θ+5cos⁡θ\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta.

Check. At θ=π3\theta = \frac{\pi}{3}, the polynomial gives 12−52+52=12=cos⁡5π3\frac12 - \frac52 + \frac52 = \frac12 = \cos\frac{5\pi}{3}.

The expansion is evidence in a proof. Do not start from the identity you have been asked to prove and merely rewrite it.

Watch out

Wrong: cos⁡3θ=cos⁡3θ\cos 3\theta = \cos^3\theta. At θ=π3\theta = \frac{\pi}{3}, the two sides would be −1-1 and 18\frac18. Right: cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta; the cross terms supply the missing pieces.

Try an even multiplier next. There is still a real term containing no cc, so expect a constant in the final polynomial.

Imaginary parts give multiple sines

The imaginary part of the same binomial expansion gives sin⁡nθ\sin n\theta. You have already done half the work: now keep the odd powers of i\mathrm{i}, then read their coefficient.

For example, i3s3=−is3\mathrm{i}^3s^3 = -\mathrm{i}s^3 contributes −s3-s^3 to the imaginary part. The imaginary part is a real number; it does not retain a factor of i\mathrm{i}.

Reading the imaginary part

For real θ\theta and a positive integer nn,

sin⁡nθ=Im⁡((cos⁡θ+isin⁡θ)n)\sin n\theta = \operatorname{Im}\bigl((\cos\theta + \mathrm{i}\sin\theta)^n\bigr)

Odd powers of i\mathrm{i} contribute, with alternating signs. The derived multiple-angle sine identities are not separately printed.

Not in the formula booklet: learn it.

Use c2=1−s2c^2 = 1-s^2 if the question wants powers of sin⁡θ\sin\theta alone. That is the same tidying decision as before, with the roles reversed.

Worked example 3: the triple-angle sine

Question. Use de Moivre's theorem to show that sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin\theta - 4\sin^3\theta.

Plan. Take the imaginary part of the cube and replace c2c^2 by 1−s21-s^2.

sin⁡3θ=Im⁡((c+is)3)de Moivre=3c2s−s3coefficient of i=3(1−s2)s−s3c2=1−s2=3s−4s3collect powers\begin{aligned} \sin 3\theta &= \operatorname{Im}\bigl((c+\mathrm{i}s)^3\bigr) && \text{de Moivre} \\ &= 3c^2s - s^3 && \text{coefficient of }\mathrm{i} \\ &= 3(1-s^2)s - s^3 && c^2=1-s^2 \\ &= 3s - 4s^3 && \text{collect powers} \end{aligned}

Why this step? The s3s^3 term is negative because i3=−i\mathrm{i}^3 = -\mathrm{i}. Its sign is decided before the trigonometric substitution.

Answer. sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin\theta - 4\sin^3\theta, as required.

Check. At θ=π6\theta = \frac{\pi}{6}, the right side is 32−12=1=sin⁡π2\frac32 - \frac12 = 1 = \sin\frac{\pi}{2}.

Change only the multiplier from 33 to 55. The odd-index terms now have signs positive, negative, positive.

Worked example 4: the quintuple-angle sine

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

Plan. Extract the imaginary terms in the fifth-power expansion and eliminate c2c^2.

sin⁡5θ=Im⁡((c+is)5)de Moivre=5c4s−10c2s3+s5i, i3, i5=5(1−s2)2s−10(1−s2)s3+s5eliminate c=5s−10s3+5s5−10s3+10s5+s5expand=16s5−20s3+5scollect powers\begin{aligned} \sin 5\theta &= \operatorname{Im}\bigl((c+\mathrm{i}s)^5\bigr) && \text{de Moivre} \\ &= 5c^4s - 10c^2s^3 + s^5 && \mathrm{i},\ \mathrm{i}^3,\ \mathrm{i}^5 \\ &= 5(1-s^2)^2s - 10(1-s^2)s^3 + s^5 && \text{eliminate }c \\ &= 5s - 10s^3 + 5s^5 - 10s^3 + 10s^5 + s^5 && \text{expand} \\ &= 16s^5 - 20s^3 + 5s && \text{collect powers} \end{aligned}

Why this step? Keeping the mixed form 5c4s−10c2s3+s55c^4s - 10c^2s^3 + s^5 visible separates the complex-number step from the algebra, making both easy to check.

Answer. sin⁡5θ=16sin⁡5θ−20sin⁡3θ+5sin⁡θ\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta.

Check. At θ=π6\theta = \frac{\pi}{6}, the result is 12−52+52=12=sin⁡5π6\frac12 - \frac52 + \frac52 = \frac12 = \sin\frac{5\pi}{6}.

For even multiples, an odd power of both sine and cosine remains. Do not force a square root into the answer unless the question specifically requires it.

Watch out

Wrong: Im⁡(3c2si−s3i)=i(3c2s−s3)\operatorname{Im}(3c^2s\mathrm{i} - s^3\mathrm{i}) = \mathrm{i}(3c^2s - s^3). Right: the imaginary part is 3c2s−s33c^2s - s^3. You equate it to sin⁡3θ\sin 3\theta, not to isin⁡3θ\mathrm{i}\sin 3\theta.

The next expansion needs only two imaginary terms.

Dividing gives multiple tangents

Divide the imaginary part by the real part to obtain tan⁡nθ\tan n\theta. Then divide every term by cos⁡nθ\cos^n\theta to express the result using t=tan⁡θt = \tan\theta.

There are two restrictions to keep in view: tt exists only when cos⁡θ≠0\cos\theta \ne 0, and tan⁡nθ\tan n\theta exists only when cos⁡nθ≠0\cos n\theta \ne 0. A zero denominator in the final rational expression records the second restriction.

Multiple-angle tangents

For real θ\theta, with t=tan⁡θt = \tan\theta defined and the displayed denominator non-zero,

tan⁡3θ=3t−t31−3t2tan⁡5θ=5t−10t3+t51−10t2+5t4\tan 3\theta = \frac{3t-t^3}{1-3t^2} \qquad \tan 5\theta = \frac{5t-10t^3+t^5}{1-10t^2+5t^4}

These identities follow by dividing the imaginary and real parts of de Moivre's expansions.

Not in the formula booklet: learn it.

Use the mixed forms from the expansion before eliminating either sine or cosine. Each term then has the same total degree, so a single division by cnc^n produces powers of tt.

Worked example 5: a triple-angle tangent

Question. Use de Moivre's theorem to show that tan⁡3θ=3t−t31−3t2\tan 3\theta = \frac{3t-t^3}{1-3t^2}, where t=tan⁡θt = \tan\theta, whenever both sides are defined.

Plan. Divide the imaginary and real parts of (c+is)3(c+\mathrm{i}s)^3, then divide top and bottom by c3c^3.

tan⁡3θ=sin⁡3θcos⁡3θdefinition of tangent=3c2s−s3c3−3cs2de Moivre’s cube=3(s/c)−(s/c)31−3(s/c)2divide by c3=3t−t31−3t2t=s/c\begin{aligned} \tan 3\theta &= \frac{\sin 3\theta}{\cos 3\theta} && \text{definition of tangent} \\ &= \frac{3c^2s-s^3}{c^3-3cs^2} && \text{de Moivre's cube} \\ &= \frac{3(s/c)-(s/c)^3}{1-3(s/c)^2} && \text{divide by }c^3 \\ &= \frac{3t-t^3}{1-3t^2} && t=s/c \end{aligned}

Why this step? Dividing by c3c^3 turns the ratios into tangent powers directly, avoiding square roots and sign choices.

Answer. The identity holds for cos⁡θ≠0\cos\theta \ne 0 and 1−3t2≠01-3t^2 \ne 0.

Check. At θ=π4\theta = \frac{\pi}{4}, t=1t=1, so the fraction is 2−2=−1=tan⁡3π4\frac{2}{-2}=-1=\tan\frac{3\pi}{4}.

Change only the multiplier from 33 to 55. Divide by the fifth power of cosine instead of the third.

Worked example 6: a quintuple-angle tangent

Question. Express tan⁡5θ\tan 5\theta as a rational function of t=tan⁡θt = \tan\theta, using de Moivre's theorem.

Plan. Use the mixed fifth-power real and imaginary parts, then divide every term by c5c^5.

tan⁡5θ=5c4s−10c2s3+s5c5−10c3s2+5cs4imaginary over real=5(s/c)−10(s/c)3+(s/c)51−10(s/c)2+5(s/c)4divide by c5=5t−10t3+t51−10t2+5t4t=tan⁡θ\begin{aligned} \tan 5\theta &= \frac{5c^4s-10c^2s^3+s^5}{c^5-10c^3s^2+5cs^4} && \text{imaginary over real} \\ &= \frac{5(s/c)-10(s/c)^3+(s/c)^5}{1-10(s/c)^2+5(s/c)^4} && \text{divide by }c^5 \\ &= \frac{5t-10t^3+t^5}{1-10t^2+5t^4} && t=\tan\theta \end{aligned}

Why this step? Both numerator and denominator come from the same expansion; using an unrelated sine identity could mix angles or powers.

Answer. tan⁡5θ=5t−10t3+t51−10t2+5t4\tan 5\theta = \frac{5t-10t^3+t^5}{1-10t^2+5t^4}, provided cos⁡θ≠0\cos\theta \ne 0 and 1−10t2+5t4≠01-10t^2+5t^4 \ne 0.

Check. At θ=π4\theta = \frac{\pi}{4}, the fraction is −4−4=1=tan⁡5π4\frac{-4}{-4}=1=\tan\frac{5\pi}{4}.

The denominator restriction is real information. For example, t=13t=\frac{1}{\sqrt{3}} makes the triple-angle denominator zero: θ=π6\theta=\frac{\pi}{6} would give 3θ=π23\theta=\frac{\pi}{2}, where tangent is undefined.

Watch out

Wrong: cancel a factor that may be zero and conclude that an identity holds at the cancelled value. Right: state the restrictions before division. A simplified fraction cannot give a value to an undefined tan⁡nθ\tan n\theta.

We can also finish the opening puzzle. With x=cos⁡π9x=\cos\frac{\pi}{9}, the triple-angle cosine gives 4x3−3x=cos⁡π3=124x^3-3x=\cos\frac{\pi}{3}=\frac12, so 8x3−6x−1=08x^3-6x-1=0. An angle calculation has produced an exact algebraic equation, without approximating the cosine.

Quick Recap

  • Expand (cos⁡θ+isin⁡θ)n(\cos\theta + \mathrm{i}\sin\theta)^n and use de Moivre to identify its two parts.
  • Real parts give cos⁡nθ\cos n\theta; imaginary parts give sin⁡nθ\sin n\theta.
  • Replace sin⁡2θ\sin^2\theta or cos⁡2θ\cos^2\theta to obtain the requested powers of one function.
  • Divide imaginary by real to obtain tan⁡nθ\tan n\theta, retaining the restrictions on both tangents.

The opening cubic follows from cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta at θ=π9\theta = \frac{\pi}{9}.