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2015 Paper 1 Q2
D: 1484.0 B: 1500.0
trigonometry exact trigonometric values triple angle formula cos 3α roots of polynomials compound angle formulae cubic equation surd form

  1. Show that \(\cos 15^\circ = \dfrac{\sqrt3 +1}{2\sqrt2}\) and find a similar expression for \(\sin 15^\circ\).
  2. Show that \(\cos \alpha\) is a root of the equation \[ 4x^3-3 x -\cos 3\alpha =0\,, \] and find the other two roots in terms of \(\cos\alpha\) and \(\sin\alpha\).
  3. Use parts (i) and (ii) to solve the equation \(y^3-3y -\sqrt2 =0\,\), giving your answers in surd form.

Solution:

  1. \begin{align*} \cos 15^{\circ} &= \cos (45^{\circ} - 30^{\circ}) \\ &= \cos 45^{\circ} \cos 30^{\circ} + \sin 45^{\circ} \sin 30^{\circ} \\ &= \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\frac{1}{2} \\ &= \frac{\sqrt{3}+1}{2\sqrt{2}} \\ \\ \sin15^{\circ} &= \sin(45^{\circ} - 30^{\circ}) \\ &= \sin45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} \\ &= \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\frac{1}{2} \\ &= \frac{\sqrt{3}-1}{2\sqrt{2}} \\ \end{align*}
  2. \begin{align*} \cos 3 \alpha &= \cos 2\alpha \cos \alpha - \sin2\alpha \sin \alpha \\ &= (2\cos^2 \alpha -1)\cos \alpha - 2 \cos \alpha \sin^2 \alpha \\ &= 2\cos^3 \alpha - \cos \alpha - 2\cos \alpha (1-\cos^2 \alpha) \\ &= 4\cos^2 \alpha - 3\cos \alpha \end{align*} Therefore if \(x = \cos \alpha\) then \(4x^3 - 3x-\cos3\alpha = 0\). \begin{align*} 0 &= 4x^3 - 3x-\cos3\alpha \\ &= 4x^3 - 3x - 4\cos^3\alpha+ 3\cos \alpha \\ &= 4(x-\cos\alpha)(x^2+x\cos\alpha+\cos^2\alpha)-3(x-\cos\alpha)\\ &= (x - \cos \alpha)(4x^2+4x\cos\alpha+4\cos^2\alpha-3) \end{align*} Therefore the other roots will be solutions to the second quadratic which are: \begin{align*} \frac{-4\cos \alpha \pm \sqrt{16\cos^2\alpha - 16(4\cos^2\alpha-3)}}{8} &= \frac{-\cos \alpha \pm \sqrt{3(1-\cos^2\alpha)}}{2} \\ &= \frac{-\cos \alpha \pm \sqrt{3} \sin \alpha}{2} \end{align*}
  3. Suppose \(y^3-3y-\sqrt{2} = 0\) then \(4\l \frac{y}{2} \r ^3-3(\frac{y}{2}) -\frac{\sqrt{2}}{2} = 0\) or alternatively, if \(x = \frac{y}{2}\), \(4x^3-3x-\cos 45^{\circ} = 0\). Therefore \(x = \cos 15^{\circ}, \frac{-\cos 15^{\circ} \pm \sqrt{3} \sin 15^{\circ}}{2}\) Therefore \(y =2\cos 15^{\circ}, -\cos 15^{\circ} \pm \sqrt{3} \sin 15^{\circ}\) or \(y = \frac{\sqrt{6}+\sqrt{2}}{2}\), \begin{align*} y &= -\frac{\sqrt{3}+1}{2\sqrt{2}} \pm \frac{3-\sqrt{3}}{2\sqrt{2}} \\ &= \frac{-4}{2\sqrt{2}}, \frac{-2\sqrt{3}}{2\sqrt{2}} \\ &= -\sqrt{2}, -\frac{\sqrt{6}-\sqrt{2}}{2} \end{align*}

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1996 Paper 2 Q5
D: 1600.0 B: 1500.0
complex numbers roots of unity quadratic trigonometric De Moivre exact trigonometric values substitution fifth roots of unity

If $$ z^{4}+z^{3}+z^{2}+z+1=0\tag{*} $$ and \(u=z+z^{-1}\), find the possible values of \(u\). Hence find the possible values of \(z\). [Do not try to simplify your answers.] Show that, if \(z\) satisfies \((*)\), then \[z^{5}-1=0.\] Hence write the solutions of \((*)\) in the form \(z=r(\cos\theta+i\sin\theta)\) for suitable real \(r\) and \(\theta\). Deduce that \[\sin\frac{2\pi}{5}=\frac{\surd(10+2\surd 5)}{4} \ \ \hbox{and}\ \ \cos\frac{2\pi}{5}=\frac{-1+\surd 5}{4}.\]

Solution: \begin{align*} && 0 &= z^4+z^3+z^2+z+1 \\ \Rightarrow && 0 &= z^2+z+1+z^{-1}+z^{-2} \tag{\(z \neq 0\)} \\ &&&= \left ( z+z^{-1} \right)^2-2 + z+z^{-1} + 1 \\ &&&= u^2+u-1 \\ \Rightarrow && u &= \frac{-1 \pm \sqrt{5}}{2} \\ \Rightarrow && z+z^{-1} &= \frac{-1 \pm \sqrt{5}}{2} \\ \Rightarrow && 0 &= z^2-\left ( \frac{-1 \pm \sqrt{5}}{2}\right)z+1 \\ \Rightarrow && z &= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\left ( \frac{-1 \pm \sqrt{5}}{2}\right)^2-4}}{2} \\ &&&= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\frac{1+5\mp2\sqrt{5}-16}{4}}}{2} \\ &&&= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\frac{-10\mp2\sqrt{5}-16}{4}}}{2} \\ &&&= \frac{-1\pm\sqrt{5}}{4} \pm i\frac{\sqrt{10\pm 2\sqrt{5}}}{4} \end{align*} Since \(z^4+z^3+z^2+z+1 = 0\) we can multiply both sides by \(z-1\) to obtain \(z^5-1 = 0\). Therefore if \(z = r(\cos \theta + i \sin \theta)\) we see that \(z^5 = 1 \Rightarrow r^5 (\cos 5 \theta + i \sin 5 \theta) = 1 \Rightarrow r = 1, 5 \theta = 2n \pi\) ie \(z = \cos \frac{2n\pi}{5} + i\sin \frac{2n \pi}{5}\). We are looking for a solution in the first quadrant, therefore \(\cos \frac{2\pi}{5} = \frac{-1 + \sqrt{5}}4\) and \(\sin \frac{2\pi}{5} = \frac{\sqrt{10+2\sqrt{5}}}{4}\)

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1995 Paper 1 Q4
D: 1484.0 B: 1500.0
complex numbers De Moivre binomial theorem trig identity sin nθ equating imaginary parts exact trigonometric values quadratic

By applying de Moivre's theorem to \(\cos5\theta+\mathrm{i}\sin5\theta,\) expanding the result using the binomial theorem, and then equating imaginary parts, show that \[ \sin5\theta=\sin\theta\left(16\cos^{4}\theta-12\cos^{2}\theta+1\right). \] Use this identity to evaluate \(\cos^{2}\frac{1}{5}\pi\), and deduce that \(\cos\frac{1}{5}\pi=\frac{1}{4}(1+\sqrt{5}).\)

Solution: \begin{align*} && (\cos \theta + i \sin \theta)^n &= \cos n \theta + i \sin n \theta \\ n = 5: && \cos 5 \theta + i \sin 5 \theta &= (\cos \theta + i \sin \theta)^5 \\ \textrm{Im}: && \sin 5 \theta &= \binom{5}{1}\cos^4 \theta \sin \theta + \binom{5}{3} \cos^2 \theta (- \sin^3 \theta) + \binom{5}{5} \sin^5 \theta \\ &&&= \sin \theta (5\cos^4 \theta-10\cos^2 \theta \sin^2 \theta+\sin^4 \theta) \\ &&&= \sin \theta (5\cos^4 \theta-10\cos^2 \theta (1-\cos^2 \theta)+(1-\cos^2 \theta)^2) \\ &&&= \sin \theta((5+10+1)\cos^4 \theta +(-10-2)\cos^2 \theta + 1) \\ &&&= \sin \theta(16\cos^4 \theta -12\cos^2 \theta + 1) \\ \end{align*} Suppose \(\theta= \frac{\pi}{5}\), then \(\sin 5 \theta = 0, \sin \theta \neq 0\), therefore if \(c = \cos \theta\) we must have \begin{align*} && 0 &= 16c^4-12c^2+1 \\ \Rightarrow && c^2 &= \frac{3 \pm \sqrt{5}}{8} \\ &&&= \frac{6\pm 2\sqrt{5}}{16} \\ &&&= \frac{(1 \pm \sqrt{5})^2}{16} \\ \Rightarrow && c &= \pm \frac{1 \pm \sqrt{5}}{4} \end{align*} Since \(c > 0\) we either have \(\cos \frac15 \pi = \frac{1+\sqrt{5}}4\) or \(\cos \frac15 \pi = \frac{\sqrt{5}-1}4\), however \(\sqrt{5}-1 < 1.5\) and so \(\frac{\sqrt{5}-1}{4} < \frac12 = \cos \frac13 \pi\) we must have \(\cos \frac15 \pi = \frac{1+\sqrt{5}}4\)

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