Home / A-Level Further Maths 9231 / Complex numbers / Multiple angles and powers of sine and cosine
Cambridge International AS & A Level · Topic 2.5
A-Level Further Maths 9231 / Topic 2.5 / AS & A Level

One binomial expansion, two kinds of identity.

De Moivre's theorem turns \(\cos n\theta\), \(\sin n\theta\) and \(\tan n\theta\) into polynomials in \(\cos\theta\), \(\sin\theta\) or \(\tan\theta\), and turns powers such as \(\cos^{5}\theta\) back into sums of multiple angles. Learn both directions, then use them to solve polynomial equations, integrate powers and find exact values like \(\cos\frac{\pi}{10}\).

Topic 2.5All complex numbers lessons →
The key idea

Write \(c = \cos\theta\), \(s = \sin\theta\). Multiple angles: by de Moivre, \(\cos n\theta + i\sin n\theta = (c + is)^{n}\); expand by the binomial theorem and equate real parts (for \(\cos n\theta\)) and imaginary parts (for \(\sin n\theta\)); \(\tan n\theta = \dfrac{\text{Im}}{\text{Re}}\), divided through by \(c^{n}\). Powers: with \(z = e^{i\theta}\), \(z + \tfrac{1}{z} = 2\cos\theta\), \(z - \tfrac{1}{z} = 2i\sin\theta\), \(z^{n} + z^{-n} = 2\cos n\theta\) and \(z^{n} - z^{-n} = 2i\sin n\theta\); expand \(\big(z \pm \tfrac{1}{z}\big)^{n}\) and pair the terms from the two ends.

x y 1 −1 −1 1 y = 16x⁵ − 20x³ + 5x with x = cos θ, y = cos 5θ y = ½ five roots x = cos θ θ = π/15, π/3, 7π/15, 11π/15, 13π/15 the curve stays between −1 and 1 for −1 ≤ x ≤ 1
Fig. 1: because cos 5θ = 16cos⁵θ - 20cos³θ + 5cos θ, the graph of y = 16x⁵ - 20x³ + 5x for -1 ≤ x ≤ 1 swings between -1 and 1 five times. A line such as y = ½ cuts it at five points, and each one is x = cos θ for a solution of cos 5θ = ½: the quintic is solved exactly with no algebra at all
Section 01

Multiple angles from real and imaginary parts.

De Moivre's theorem gives \(\cos n\theta + i\sin n\theta = (\cos\theta + i\sin\theta)^{n}\). The left side is already split into a real part and an imaginary part. Expand the right side with the binomial theorem, using \(i^{2} = -1\), \(i^{3} = -i\), \(i^{4} = 1\), and equate the two parts. Writing \(c = \cos\theta\) and \(s = \sin\theta\) keeps the working short.

Term of \((c + is)^{n}\)Power of \(i\)Goes to
\(c^{n}\), \(\binom{n}{2}c^{n-2}(is)^{2}\), \(\binom{n}{4}c^{n-4}(is)^{4}\), ...even: \(i^{2} = -1\), \(i^{4} = 1\)real part, signs \(+, -, +, \ldots\)
\(\binom{n}{1}c^{n-1}(is)\), \(\binom{n}{3}c^{n-3}(is)^{3}\), ...odd: \(i\), \(i^{3} = -i\)imaginary part, signs \(+, -, +, \ldots\)

So the real part uses the even-numbered binomial coefficients and the imaginary part the odd ones, each with alternating signs. To write \(\cos n\theta\) in powers of \(\cos\theta\) only, replace every \(s^{2}\) by \(1 - c^{2}\); for \(\sin n\theta\) in powers of \(\sin\theta\) only, replace every \(c^{2}\) by \(1 - s^{2}\). This works neatly when the unwanted function appears only to even powers: \(\cos n\theta\) always, and \(\sin n\theta\) when \(n\) is odd.

Worked example cos 4θ in powers of cos θ
Use de Moivre's theorem to show that \(\cos 4\theta = 8\cos^{4}\theta - 8\cos^{2}\theta + 1\).
  1. By de Moivre, \(\cos 4\theta + i\sin 4\theta = (c + is)^{4}\). Expand: \[\begin{aligned}(c + is)^{4} &= c^{4} + 4c^{3}(is) + 6c^{2}(is)^{2} \\ &\quad + 4c(is)^{3} + (is)^{4} \\ &= c^{4} + 4ic^{3}s - 6c^{2}s^{2} \\ &\quad - 4ics^{3} + s^{4}.\end{aligned}\]
  2. Equate real parts: \(\cos 4\theta = c^{4} - 6c^{2}s^{2} + s^{4}\).
  3. Replace \(s^{2}\) by \(1 - c^{2}\): \[\begin{aligned}\cos 4\theta &= c^{4} - 6c^{2}(1 - c^{2}) + (1 - c^{2})^{2} \\ &= c^{4} - 6c^{2} + 6c^{4} + 1 - 2c^{2} + c^{4} \\ &= 8c^{4} - 8c^{2} + 1.\end{aligned}\]
  4. Check at \(\theta = 0\): both sides equal \(1\). At \(\theta = \frac{\pi}{4}\): \(c^{2} = \frac{1}{2}\), so \(8 \times \frac{1}{4} - 8 \times \frac{1}{2} + 1 = -1 = \cos\pi\).
Answer: \(\cos 4\theta = 8\cos^{4}\theta - 8\cos^{2}\theta + 1\), as required
Worked example sin 5θ in powers of sin θ
Express \(\sin 5\theta\) in terms of \(\sin\theta\) only.
  1. \(\sin 5\theta\) is the imaginary part of \((c + is)^{5}\). The odd terms are \(\binom{5}{1}c^{4}(is)\), \(\binom{5}{3}c^{2}(is)^{3}\) and \((is)^{5}\), with \(i^{3} = -i\) and \(i^{5} = i\).
  2. So \(\sin 5\theta = 5c^{4}s - 10c^{2}s^{3} + s^{5}\).
  3. \(c\) appears only as \(c^{2}\) and \(c^{4}\), so put \(c^{2} = 1 - s^{2}\): \[\begin{aligned}\sin 5\theta &= 5s(1 - s^{2})^{2} - 10s^{3}(1 - s^{2}) + s^{5} \\ &= 5s - 10s^{3} + 5s^{5} \\ &\quad - 10s^{3} + 10s^{5} + s^{5} \\ &= 16s^{5} - 20s^{3} + 5s.\end{aligned}\]
Answer: \(\sin 5\theta = 16\sin^{5}\theta - 20\sin^{3}\theta + 5\sin\theta\)
Your turn
Use de Moivre's theorem to show that \(\sin 4\theta = \sin\theta\,(8\cos^{3}\theta - 4\cos\theta)\).
Examiner trap

Leaving out the powers of \(i\) is the classic slip: \((is)^{2} = -s^{2}\), not \(s^{2}\), and \((is)^{3} = -is^{3}\). Without them every sign after the first term is wrong, and \(\cos 4\theta\) comes out as \(c^{4} + 6c^{2}s^{2} + s^{4} = (c^{2} + s^{2})^{2} = 1\), which is clearly false. Also keep each binomial coefficient: \(\binom{5}{3} = 10\), not \(5\).

Section 02

tan nθ, and powers back to multiple angles.

For \(\tan n\theta\), divide the imaginary part by the real part, then divide every term top and bottom by \(c^{n}\), so that each \(\dfrac{s}{c}\) becomes \(t = \tan\theta\). Going the other way, a power such as \(\cos^{5}\theta\) is written as a sum of \(\cos 5\theta\), \(\cos 3\theta\), \(\cos\theta\) by expanding \(\big(z + \tfrac{1}{z}\big)^{5} = (2\cos\theta)^{5}\), where \(z = \cos\theta + i\sin\theta\).

The pairing step is what makes the powers method work. In the expansion of \(\big(z + \tfrac{1}{z}\big)^{n}\) the first and last terms are \(z^{n}\) and \(z^{-n}\), the second and second-last are \(n z^{n-2}\) and \(n z^{-(n-2)}\), and so on, with equal coefficients. Each pair is \(z^{k} + z^{-k} = 2\cos k\theta\). For sines use \(\big(z - \tfrac{1}{z}\big)^{n} = (2i\sin\theta)^{n}\), whose pairs are \(z^{k} - z^{-k} = 2i\sin k\theta\) or \(z^{k} + z^{-k} = 2\cos k\theta\), depending on the signs.

\(n\)\((2i\sin\theta)^{n}\)Pairs give
2, 6, 10, ...\(-2^{n}\sin^{n}\theta\)cosines
4, 8, ...\(+2^{n}\sin^{n}\theta\)cosines
odd\(\pm 2^{n}i\sin^{n}\theta\)sines
Worked example tan 3θ in powers of tan θ
Use de Moivre's theorem to show that \(\tan 3\theta = \dfrac{3\tan\theta - \tan^{3}\theta}{1 - 3\tan^{2}\theta}\).
  1. \((c + is)^{3} = c^{3} + 3ic^{2}s - 3cs^{2} - is^{3}\), so \(\cos 3\theta = c^{3} - 3cs^{2}\) and \(\sin 3\theta = 3c^{2}s - s^{3}\).
  2. Divide: \(\tan 3\theta = \dfrac{3c^{2}s - s^{3}}{c^{3} - 3cs^{2}}\).
  3. Divide every term top and bottom by \(c^{3}\), using \(\dfrac{s}{c} = t\): \[\tan 3\theta = \frac{3t - t^{3}}{1 - 3t^{2}}.\]
Answer: \(\tan 3\theta = \dfrac{3\tan\theta - \tan^{3}\theta}{1 - 3\tan^{2}\theta}\), as required
Worked example cos⁵θ in multiple angles
Express \(\cos^{5}\theta\) in the form \(a\cos 5\theta + b\cos 3\theta + c\cos\theta\).
  1. With \(z = \cos\theta + i\sin\theta\), \(z + \tfrac{1}{z} = 2\cos\theta\), so \(32\cos^{5}\theta = \big(z + \tfrac{1}{z}\big)^{5}\).
  2. Expand and pair from the ends: \[\begin{aligned}\Big(z + \frac{1}{z}\Big)^{5} &= z^{5} + 5z^{3} + 10z \\ &\quad + \frac{10}{z} + \frac{5}{z^{3}} + \frac{1}{z^{5}} \\ &= \Big(z^{5} + \frac{1}{z^{5}}\Big) + 5\Big(z^{3} + \frac{1}{z^{3}}\Big) \\ &\quad + 10\Big(z + \frac{1}{z}\Big).\end{aligned}\]
  3. Each bracket is \(2\cos k\theta\): \(32\cos^{5}\theta = 2\cos 5\theta + 10\cos 3\theta + 20\cos\theta\).
  4. Divide by \(32\): \(\cos^{5}\theta = \tfrac{1}{16}(\cos 5\theta + 5\cos 3\theta + 10\cos\theta)\). Check at \(\theta = 0\): \(\tfrac{1}{16}(1 + 5 + 10) = 1\).
Answer: \(\cos^{5}\theta = \tfrac{1}{16}\cos 5\theta + \tfrac{5}{16}\cos 3\theta + \tfrac{5}{8}\cos\theta\)
Your turn
Show that \(\sin^{4}\theta = \tfrac{1}{8}(\cos 4\theta - 4\cos 2\theta + 3)\).
Examiner trap

Two factors are easy to lose. First, \(z + \tfrac{1}{z}\) is \(2\cos\theta\), so the left side is \(2^{n}\cos^{n}\theta\), not \(\cos^{n}\theta\). Second, each pair \(z^{k} + z^{-k}\) is \(2\cos k\theta\), not \(\cos k\theta\). The middle constant term of an even power (the \(6\) in the fourth power) has no partner and is NOT doubled. A quick check at \(\theta = 0\) catches most of these errors.

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Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

Equating real parts in \((\cos\theta + i\sin\theta)^{4}\) gives \(\cos 4\theta =\)

Quick check+10 XP

Which of these is \(\tan 3\theta\) in terms of \(t = \tan\theta\)?

Quick check+10 XP

With \(z = \cos\theta + i\sin\theta\), \(\big(z + \tfrac{1}{z}\big)^{4} = z^{4} + 4z^{2} + 6 + 4z^{-2} + z^{-4}\). Hence \(16\cos^{4}\theta =\)

Quick check+10 XP

With \(z = \cos\theta + i\sin\theta\), \(\big(z - \tfrac{1}{z}\big)^{6}\) is equal to

Section 03

Putting the results to work.

These identities are rarely the end of a question. A polynomial such as \(16x^{5} - 20x^{3} + 5x\) is \(\cos 5\theta\) in disguise when \(x = \cos\theta\), so an equation in \(x\) becomes an easy trigonometric equation. A power such as \(\sin^{4}\theta\), which cannot be integrated directly, becomes a sum of cosines that can. And setting \(\cos 5\theta = 0\) at a known angle gives a polynomial whose roots are exact values.

Three habits make these questions reliable:

TaskMethodWatch for
Solve a polynomial in \(\cos\theta\)recognise \(\cos n\theta\); solve for \(n\theta\) over the stretched range\(0 \le \theta \le \pi\) means \(0 \le 5\theta \le 5\pi\)
Integrate \(\sin^{n}\theta\) or \(\cos^{n}\theta\) (\(n\) even)use the multiple-angle form; \(\int \cos k\theta \, d\theta = \frac{1}{k}\sin k\theta\)the constant term gives the \(\pi\)
Exact valueschoose \(\theta\) so \(\cos n\theta\) or \(\sin n\theta\) is known; solve the polynomialpick the root that matches the sign and size of the angle
Worked example Recognise cos 5θ
Given \(\cos 5\theta = 16\cos^{5}\theta - 20\cos^{3}\theta + 5\cos\theta\), solve \(32\cos^{5}\theta - 40\cos^{3}\theta + 10\cos\theta = 1\) for \(0 \le \theta \le \pi\).
  1. The left side is \(2(16c^{5} - 20c^{3} + 5c) = 2\cos 5\theta\), so the equation is \(\cos 5\theta = \tfrac{1}{2}\).
  2. \(0 \le \theta \le \pi\) gives \(0 \le 5\theta \le 5\pi\). In that range \(\cos 5\theta = \tfrac{1}{2}\) when \(5\theta = \tfrac{\pi}{3}, \tfrac{5\pi}{3}, \tfrac{7\pi}{3}, \tfrac{11\pi}{3}, \tfrac{13\pi}{3}\).
  3. Divide by \(5\). A quintic has at most five roots, and Fig. 1 shows the line \(y = \frac{1}{2}\) meeting the curve five times, so there are no more.
Answer: \(\theta = \tfrac{\pi}{15}, \tfrac{\pi}{3}, \tfrac{7\pi}{15}, \tfrac{11\pi}{15}, \tfrac{13\pi}{15}\)
Worked example An exact value from a polynomial
By considering \(\cos 5\theta = 0\), show that \(\cos^{2}\tfrac{\pi}{10} = \dfrac{5 + \sqrt{5}}{8}\).
  1. At \(\theta = \frac{\pi}{10}\), \(5\theta = \frac{\pi}{2}\) and \(\cos 5\theta = 0\), so \(16c^{5} - 20c^{3} + 5c = 0\) with \(c = \cos\frac{\pi}{10} \ne 0\).
  2. Divide by \(c\): \(16c^{4} - 20c^{2} + 5 = 0\), a quadratic in \(c^{2}\).
  3. \[c^{2} = \frac{20 \pm \sqrt{400 - 320}}{32} = \frac{20 \pm 4\sqrt{5}}{32} = \frac{5 \pm \sqrt{5}}{8}.\]
  4. The two roots are \(\cos^{2}\frac{\pi}{10}\) and \(\cos^{2}\frac{3\pi}{10}\) (also a solution of \(\cos 5\theta = 0\)). Since \(\frac{\pi}{10} \lt \frac{3\pi}{10}\) and cosine decreases on \(\big[0, \frac{\pi}{2}\big]\), \(\cos^{2}\frac{\pi}{10}\) is the larger root.
Answer: \(\cos^{2}\tfrac{\pi}{10} = \dfrac{5 + \sqrt{5}}{8}\), so \(\cos\tfrac{\pi}{10} = \sqrt{\dfrac{5 + \sqrt{5}}{8}} \approx 0.9511\)
Your turn
Using \(\sin^{4}\theta = \tfrac{1}{8}(\cos 4\theta - 4\cos 2\theta + 3)\), find the exact value of \(\displaystyle\int_{0}^{\pi/2} \sin^{4}\theta \, d\theta\).
Examiner trap

Solving \(\cos 5\theta = \frac{1}{2}\) only over \(0 \le 5\theta \le \pi\) gives one answer instead of five: stretch the range by the same factor first, then divide back. In exact-value questions, do not just take the '+' root out of habit: decide which angle each root belongs to, using the sign and size of the cosine or sine.

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Using \(\tan 3\theta = \dfrac{3t - t^{3}}{1 - 3t^{2}}\) with \(t = \tan\theta\), the roots of \(t^{3} - 3t^{2} - 3t + 1 = 0\) are

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Given \(\sin^{4}\theta = \tfrac{1}{8}(\cos 4\theta - 4\cos 2\theta + 3)\), find \(\displaystyle\int_{0}^{\pi/4} \sin^{4}\theta \, d\theta\).

Exam style+20 XP

\(\theta = \frac{\pi}{10}\) satisfies \(\sin 5\theta = 1\), and \(16s^{5} - 20s^{3} + 5s - 1 = (s - 1)(4s^{2} + 2s - 1)^{2}\), where \(s = \sin\theta\). Find the exact value of \(\sin\frac{\pi}{10}\).

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Express \(\sin^{5}\theta\) in terms of \(\sin 5\theta\), \(\sin 3\theta\) and \(\sin\theta\).

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Multiple angles expanded, powers paired, polynomials solved.

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Six original Cambridge-style questions
Model derivations of \(\sin 3\theta\), \(\cos 6\theta\) and \(\tan 5\theta\), \(\cos^{6}\theta\) and \(\sin^{2}\theta\cos^{4}\theta\) in multiple angles with exact integrals, and the quintic for \(\cos 5\theta = -1\) solved to give \(\cos\frac{\pi}{5}\).