AS & A Level · Practice questions · Multiple angles and powers of sine and cosine

Expand, pair, then use it.

Six original Cambridge-style questions on de Moivre's theorem: multiple angles in powers of \(\cos\theta\), \(\sin\theta\) and \(\tan\theta\), powers in multiple angles, exact integrals, and polynomial equations solved by spotting \(\cos n\theta\) or \(\tan n\theta\).

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Binomial out, pairs back.

Multiple angle: expand \((c + is)^{n}\); real part is \(\cos n\theta\), imaginary part is \(\sin n\theta\). Track \(i^{2} = -1\), \(i^{3} = -i\).
One function only: replace \(s^{2}\) by \(1 - c^{2}\) (or \(c^{2}\) by \(1 - s^{2}\)).
tan: \(\tan n\theta = \dfrac{\text{Im}}{\text{Re}}\), then divide every term by \(c^{n}\).
Powers: \(\big(z + \tfrac{1}{z}\big)^{n} = 2^{n}\cos^{n}\theta\), \(\big(z - \tfrac{1}{z}\big)^{n} = (2i)^{n}\sin^{n}\theta\); pair \(z^{k} \pm z^{-k}\) into \(2\cos k\theta\) or \(2i\sin k\theta\).
Use it: spot \(\cos n\theta\) in a polynomial, stretch the range by \(n\), and check each root against the sign and size of the angle.

Recall

[3 marks]

Use de Moivre's theorem to show that \(\sin 3\theta = 3\sin\theta - 4\sin^{3}\theta\).

\((c + is)^{3} = c^{3} + 3ic^{2}s - 3cs^{2} - is^{3}\). ✓
Equate imaginary parts: \(\sin 3\theta = 3c^{2}s - s^{3}\). ✓
Replace \(c^{2}\) by \(1 - s^{2}\): \(\sin 3\theta = 3s(1 - s^{2}) - s^{3} = 3s - 4s^{3}\), as required. ✓

Calculation

[5 marks]

Use de Moivre's theorem to show that \(\cos 6\theta = 32\cos^{6}\theta - 48\cos^{4}\theta + 18\cos^{2}\theta - 1\).

The real terms of \((c + is)^{6}\) are the even ones: \(c^{6} + 15c^{4}(is)^{2} + 15c^{2}(is)^{4} + (is)^{6}\), so \(\cos 6\theta = c^{6} - 15c^{4}s^{2} + 15c^{2}s^{4} - s^{6}\). ✓
Put \(s^{2} = 1 - c^{2}\), so \(s^{4} = 1 - 2c^{2} + c^{4}\) and \(s^{6} = 1 - 3c^{2} + 3c^{4} - c^{6}\). ✓
\[\begin{aligned}\cos 6\theta &= c^{6} - 15c^{4} + 15c^{6} \\ &\quad + 15c^{2} - 30c^{4} + 15c^{6} \\ &\quad - 1 + 3c^{2} - 3c^{4} + c^{6}.\end{aligned}\] ✓
Collect: \(c^{6}\): \(1 + 15 + 15 + 1 = 32\); \(c^{4}\): \(-15 - 30 - 3 = -48\); \(c^{2}\): \(15 + 3 = 18\); constant \(-1\). ✓
So \(\cos 6\theta = 32c^{6} - 48c^{4} + 18c^{2} - 1\). Check at \(\theta = 0\): \(32 - 48 + 18 - 1 = 1\). ✓

Analysis

[6 marks]

(a) Use de Moivre's theorem to show that \[\tan 5\theta = \frac{5t - 10t^{3} + t^{5}}{1 - 10t^{2} + 5t^{4}}, \quad t = \tan\theta.\]

(b) Hence show that the roots of \(t^{4} - 10t^{2} + 5 = 0\) are \(\tan\frac{k\pi}{5}\) for \(k = 1, 2, 3, 4\), and deduce the exact value of \(\tan^{2}\frac{\pi}{5}\).

(a) Real part of \((c + is)^{5}\): \(c^{5} - 10c^{3}s^{2} + 5cs^{4}\). Imaginary part: \(5c^{4}s - 10c^{2}s^{3} + s^{5}\). ✓
\(\tan 5\theta = \dfrac{5c^{4}s - 10c^{2}s^{3} + s^{5}}{c^{5} - 10c^{3}s^{2} + 5cs^{4}}\); divide every term by \(c^{5}\) to get the result. ✓
(b) \(\tan 5\theta = 0\) when \(5\theta = k\pi\), so \(\theta = \frac{k\pi}{5}\). The numerator must vanish: \(t(t^{4} - 10t^{2} + 5) = 0\). ✓
\(k = 0\) gives \(t = 0\); \(k = 1, 2, 3, 4\) give four different non-zero values \(\tan\frac{k\pi}{5}\), which must be the four roots of the quartic. ✓
Put \(u = t^{2}\): \(u^{2} - 10u + 5 = 0\), so \(u = 5 \pm 2\sqrt{5}\). ✓
\(\tan^{2}\frac{\pi}{5} \lt \tan^{2}\frac{2\pi}{5}\) because \(\tan\) increases on \(\big(0, \frac{\pi}{2}\big)\), so \(\tan^{2}\frac{\pi}{5} = 5 - 2\sqrt{5}\) (\(\approx 0.528\)). ✓

Calculation

[7 marks]

(a) Show that \(\cos^{6}\theta = \tfrac{1}{32}(\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10)\).

(b) Hence find the exact value of \(\displaystyle\int_{0}^{\pi/4} \cos^{6}\theta \, d\theta\).

(a) \(\big(z + \tfrac{1}{z}\big)^{6} = 64\cos^{6}\theta\). The expansion is \(z^{6} + 6z^{4} + 15z^{2} + 20 + 15z^{-2} + 6z^{-4} + z^{-6}\). ✓
Pair: \((z^{6} + z^{-6}) + 6(z^{4} + z^{-4}) + 15(z^{2} + z^{-2}) + 20 = 2\cos 6\theta + 12\cos 4\theta + 30\cos 2\theta + 20\). ✓
Divide by \(64\): \(\cos^{6}\theta = \tfrac{1}{32}(\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10)\). ✓
(b) \[\begin{aligned}\int_{0}^{\pi/4} \cos^{6}\theta \, d\theta &= \frac{1}{32}\Big[\frac{\sin 6\theta}{6} + \frac{3\sin 4\theta}{2} \\ &\qquad + \frac{15\sin 2\theta}{2} + 10\theta\Big]_{0}^{\pi/4}.\end{aligned}\] ✓
At \(\frac{\pi}{4}\): \(\sin\frac{3\pi}{2} = -1\), \(\sin\pi = 0\), \(\sin\frac{\pi}{2} = 1\). All terms are \(0\) at \(0\). ✓
\(\tfrac{1}{32}\big(-\tfrac{1}{6} + \tfrac{15}{2} + \tfrac{5\pi}{2}\big) = \tfrac{1}{32}\big(\tfrac{22}{3} + \tfrac{5\pi}{2}\big) = \tfrac{11}{48} + \tfrac{5\pi}{64}\) (\(\approx 0.4746\)). ✓

Analysis

[6 marks]

By considering \(\big(z + \tfrac{1}{z}\big)^{4}\big(z - \tfrac{1}{z}\big)^{2}\), where \(z = \cos\theta + i\sin\theta\), show that \[\sin^{2}\theta\cos^{4}\theta = \tfrac{1}{32}(2 + \cos 2\theta - 2\cos 4\theta - \cos 6\theta),\] and hence find \(\displaystyle\int_{0}^{\pi/2} \sin^{2}\theta\cos^{4}\theta \, d\theta\).

The product is \((2\cos\theta)^{4}(2i\sin\theta)^{2} = 16\cos^{4}\theta \times (-4\sin^{2}\theta) = -64\sin^{2}\theta\cos^{4}\theta\). ✓
Group as \(\big(z + \tfrac{1}{z}\big)^{2}\Big[\big(z + \tfrac{1}{z}\big)\big(z - \tfrac{1}{z}\big)\Big]^{2} = (z^{2} + 2 + z^{-2})(z^{4} - 2 + z^{-4})\). ✓
Multiply out: \[\begin{aligned}&z^{6} - 2z^{2} + z^{-2} + 2z^{4} - 4 \\ &\quad + 2z^{-4} + z^{2} - 2z^{-2} + z^{-6}.\end{aligned}\] ✓
Pair: \((z^{6} + z^{-6}) + 2(z^{4} + z^{-4}) - (z^{2} + z^{-2}) - 4 = 2\cos 6\theta + 4\cos 4\theta - 2\cos 2\theta - 4\). ✓
Divide by \(-64\): \(\sin^{2}\theta\cos^{4}\theta = \tfrac{1}{32}(2 + \cos 2\theta - 2\cos 4\theta - \cos 6\theta)\). ✓
On \(\big[0, \frac{\pi}{2}\big]\) each \(\cos 2k\theta\) integrates to \(\frac{1}{2k}\sin 2k\theta\), which is \(0\) at both limits, so the integral is \(\tfrac{1}{32} \times 2 \times \tfrac{\pi}{2} = \tfrac{\pi}{32}\). ✓

Proof

[8 marks]

(a) Use de Moivre's theorem to show that \(\cos 5\theta = 16\cos^{5}\theta - 20\cos^{3}\theta + 5\cos\theta\).

(b) Hence find all values of \(\theta\) with \(0 \le \theta \le \pi\) for which \(16x^{5} - 20x^{3} + 5x + 1 = 0\), where \(x = \cos\theta\).

(c) Given that \(16x^{5} - 20x^{3} + 5x + 1 = (x + 1)(4x^{2} - 2x - 1)^{2}\), deduce the exact values of \(\cos\frac{\pi}{5}\) and \(\cos\frac{3\pi}{5}\).

(a) Real part of \((c + is)^{5}\): \(\cos 5\theta = c^{5} - 10c^{3}s^{2} + 5cs^{4}\). ✓
With \(s^{2} = 1 - c^{2}\): \[\begin{aligned}\cos 5\theta &= c^{5} - 10c^{3}(1 - c^{2}) + 5c(1 - c^{2})^{2} \\ &= c^{5} - 10c^{3} + 10c^{5} \\ &\quad + 5c - 10c^{3} + 5c^{5} \\ &= 16c^{5} - 20c^{3} + 5c.\end{aligned}\] ✓
(b) The equation is \(\cos 5\theta + 1 = 0\), so \(\cos 5\theta = -1\) with \(0 \le 5\theta \le 5\pi\): \(5\theta = \pi, 3\pi, 5\pi\). ✓
So \(\theta = \frac{\pi}{5}, \frac{3\pi}{5}, \pi\). ✓
(c) \(\theta = \pi\) gives the factor \(x + 1\). The other two values, \(\cos\frac{\pi}{5}\) and \(\cos\frac{3\pi}{5}\), are the roots of \(4x^{2} - 2x - 1 = 0\) (each a double root of the quintic). ✓
\(x = \dfrac{2 \pm \sqrt{4 + 16}}{8} = \dfrac{1 \pm \sqrt{5}}{4}\). ✓
\(\cos\frac{\pi}{5} \gt 0\) and \(\cos\frac{3\pi}{5} \lt 0\), so \(\cos\frac{\pi}{5} = \dfrac{1 + \sqrt{5}}{4}\) and \(\cos\frac{3\pi}{5} = \dfrac{1 - \sqrt{5}}{4}\). ✓

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