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2011 Paper 2 Q4
D: 1600.0 B: 1470.8
trigonometry trig identity sin nθ quadratic in sin exact values sin 18° double angle formula product-to-sum

  1. Find all the values of \(\theta\), in the range \(0^\circ < \theta < 180^\circ\), for which \(\cos\theta=\sin 4\theta\). Hence show that \[ \sin 18^\circ = \frac14\left( \sqrt 5 -1\right). \]
  2. Given that \[ 4\sin^2 x + 1 = 4\sin^2 2x \,, \] find all possible values of \(\sin x\,\), giving your answers in the form \(p+q\sqrt5\) where \(p\) and \(q\) are rational numbers.
  3. Hence find two values of \(\alpha\) with \(0^\circ < \alpha < 90^\circ\) for which \[ \sin^23\alpha + \sin^25\alpha = \sin^2 6\alpha\,. \]

Solution:

  1. Note that \(\cos \theta = \sin (90^\circ - \theta)\) so \begin{align*} && \sin(90^\circ - \theta) &= \sin 4 \theta\\ && 90^\circ - \theta &= 4\theta +360^{\circ}k \\ && 90^\circ + \theta &= 4\theta +360^{\circ}k \\ \Rightarrow && 5\theta &= 90^\circ, 450^\circ, 810^\circ, \cdots \\ && 3 \theta &= 90^\circ, 450^\circ, \cdots \\ \Rightarrow && \theta &= 18^\circ, 90^\circ, 162^\circ, \ldots \\ && \theta &= 30^\circ, 150^\circ, \ldots \end{align*} Therefore \(\theta = 8^\circ, 30^\circ, 90^\circ, 150^\circ, 162^\circ\). Note also that: \begin{align*} && 0 &= \sin 4 \theta - \cos \theta \\ &&&= 2 \sin 2 \theta \cos 2 \theta- \cos \theta \\ &&&= 4 \sin \theta \cos \theta \cos 2 \theta - \cos \theta \\ &&&= \cos \theta \left (4 \sin \theta (1- 2\sin^2 \theta) - 1 \right) \\ &&&= \cos \theta \left (-8\sin^3 \theta +4\sin \theta - 1 \right) \\ &&&= \cos \theta (1 - 2 \sin \theta)(4 \sin^2 \theta+2\sin \theta -1)\\ \cos \theta = 0: && \theta &= 90^\circ \\ \sin \theta = \frac12: && \theta &= 30^{\circ} \\ && \theta &= \sin^{-1} \left ( \frac{-1\pm \sqrt5}{4} \right) \end{align*} Therefore \(\sin 18^{\circ} = \frac{\pm \sqrt{5}-1}{4}\), but since \(\sin 18^{\circ} > 0\) it must be the positive version.
  2. \(\,\) \begin{align*} && 4 \sin^2 x + 1 &= 4 \sin^2 2 x \\ &&&= 16 \sin^2 x \cos^2 x \\ &&&= 16 \sin^2 x (1- \sin^2 x) \\ \Rightarrow && 0 &= 16y^2 -12y+1 \\ \Rightarrow && \sin^2 x &= \frac{3\pm \sqrt5}{8} \\ &&&= \left ( \frac{1 \pm \sqrt5}{4} \right)^2 \\ \Rightarrow && \sin x &= \pm \frac{1 \pm \sqrt{5}}{4} \end{align*}
  3. \(\,\) \begin{align*} && \sin^2 x + \frac1{2^2} &= \sin^2 2x \end{align*} So if we can have \(\sin 5x = \pm \frac12\) and \(\sin 3x = \pm \frac{1 \pm \sqrt5}{4}\) then we are good, ie \begin{align*} && 5x &= 30^{\circ}, 150^{\circ}, 210^{\circ}, 330^{\circ}, 390^{\circ}, \cdots \\ \Rightarrow && x &= 6^{\circ}, 30^{\circ}, 42^{\circ}, 66^{\circ}, 78^{\circ} \\ \Rightarrow && 3x &= \boxed{18^{\circ}}, \cancel{90^{\circ}}, \boxed{126^{\circ}}, \boxed{198^{\circ}}, \boxed{78^{\circ}} \end{align*} So our solutions are \(x = 6^{\circ}, 42^{\circ}, 66^{\circ}, 78^{\circ}\) although it's interesting to note that \(x = 45^{\circ}\) is another solution

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2009 Paper 2 Q3
D: 1600.0 B: 1500.0
trigonometric identities tan half-angle exact values rationalising denominator sec and tan compound angle formula surds

Prove that \[ \tan \left ( \tfrac14 \pi -\tfrac12 x \right)\equiv \sec x -\tan x\,. \tag{\(*\)} \]

  1. Use \((*)\) to find the value of \(\tan\frac18\pi\,\). Hence show that \[ \tan \tfrac{11}{24} \pi = \frac{\sqrt3 + \sqrt2 -1}{\sqrt3 -\sqrt6+1}\;. \]
  2. Show that \[ \frac{\sqrt3 + \sqrt2 -1}{\sqrt3 -\sqrt6+1}= 2+\sqrt2+\sqrt3+\sqrt6\,. \]
  3. Use \((*)\) to show that \[ \tan \tfrac1{48}\pi = \sqrt{16+10\sqrt2+8\sqrt3 +6\sqrt6 \ }-2-\sqrt2-\sqrt3-\sqrt6\,. \]

Solution: \begin{align*} && \tan \left ( \tfrac14 \pi -\tfrac12 x \right) &\equiv \frac{\tan \tfrac{\pi}{4} - \tan \tfrac12 x}{1 + \tan \tfrac{\pi}{4} \tan \tfrac12 x} \\ &&&= \frac{1-\tan \tfrac12 x}{1+\tan \tfrac12 x} \\ \\ && \sec x - \tan x &= \frac{1+t^2}{1-t^2} - \frac{2t}{1-t^2} \\ &&&= \frac{(1-t)^2}{(1-t)(1+t)} \\ &&&= \frac{1-t}{1+t} \end{align*} Therefore both sides are equal to the same thing.

  1. \(\tan \tfrac18 \pi = \tan(\tfrac14 \pi - \tfrac12 \tfrac14\pi) = \sec \tfrac14 \pi - \tan \tfrac14 \pi = \sqrt{2} - 1\) \begin{align*} && \tan \tfrac{11}{24} \pi &= \tan (\tfrac13 \pi +\tfrac18 \pi) \\ &&&= \frac{\tan \tfrac13 \pi +\tan \tfrac18 \pi}{1-\tan \tfrac13 \pi \tan \tfrac18 \pi} \\ &&&= \frac{\sqrt{3} + \sqrt{2} - 1}{1 - \sqrt{3}(\sqrt{2}-1)} \\ &&&= \frac{\sqrt{3} + \sqrt{2} - 1}{1 +\sqrt{3}-\sqrt{6}} \\ \end{align*}
  2. \(\,\) \begin{align*} && (\sqrt{3}-\sqrt{6}+1)(2+\sqrt{2}+\sqrt{3}+\sqrt{6}) &= (2\sqrt{3}+\sqrt{6}+3+3\sqrt{2}) + \\ &&&\quad+(-2\sqrt{6}-2\sqrt{3}-3\sqrt{2}-6) + \\ &&&\quad+(2+\sqrt{2}+\sqrt{3}+\sqrt{6}) \\ &&&= \sqrt{3}+\sqrt{2}-1 \end{align*}
  3. \(\,\) \begin{align*} && \tan \tfrac{1}{48} \pi &= \tan (\tfrac14\pi - \tfrac{11}{48} \pi) \\ &&&= \sec \tfrac{11}{24} \pi - \tan \tfrac{11}{24} \pi \\ &&&= \sqrt{1+\tan^2 \tfrac{11}{24}\pi} - \tan \tfrac{11}{24} \pi \\ &&&= \sqrt{1 + (2+\sqrt{2}+\sqrt{3}+\sqrt{6})^2} - (2+\sqrt{2}+\sqrt{3}+\sqrt{6}) \\ &&&= \sqrt{16+10\sqrt{2} + 8\sqrt{3}+6\sqrt{6}} - 2 - \sqrt2 - \sqrt3-\sqrt6 \end{align*}

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2005 Paper 1 Q4
D: 1500.0 B: 1500.0
trigonometry double angle formula triple angle formula tan 3θ identity exact values quadrant determination cubic equation

  1. Given that \(\displaystyle \cos \theta = \frac35\) and that \(\displaystyle \frac{3\pi }{ 2} \le \theta \le 2\pi\), show that \(\displaystyle \sin 2 \theta = -\frac{24}{25}\), and evaluate \(\cos 3 \theta\).
  2. Prove the identity \(\displaystyle \tan 3\theta \equiv \frac {3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}\). Hence evaluate \(\tan \theta\), given that \(\displaystyle \tan 3\theta = \frac{11}{ 2}\) and that \(\displaystyle \frac{\pi}{ 4} \le \theta \le \frac{\pi}{2}\).

Solution:

  1. Since \(\cos^2 \theta + \sin^2 \theta \equiv 1\), \(\sin \theta = \pm \frac45\) and since \(\displaystyle \frac{3\pi }{ 2} \le \theta \le 2\pi\) it must be the case that \(\sin\) is negative, ie \(\sin \theta = -\frac45\). Therefore \(\sin 2 \theta = 2 \sin \theta \cos \theta = 2 \cdot \frac35 \cdot (-\frac45) = -\frac{24}{25}\). \begin{align*} \cos 3 \theta &= \cos 2 \theta \cos \theta - \sin 2\theta \sin \theta \\ &= (\cos^2 \theta - \sin^2 \theta) \cos \theta - \sin 2 \theta \sin \theta \\ &= (\frac{9}{25} - \frac{16}{25}) \frac35 + \frac{24}{25} \cdot (-\frac{4}{5}) \\ &= -\frac{21}{125} - \frac{96}{125} \\ &= -\frac{117}{125} \end{align*}
  2. \begin{align*} \tan 3 \theta &\equiv \frac{\tan 2 \theta + \tan \theta}{1 - \tan 2 \theta \tan \theta} \\ &\equiv \frac{\frac{2 \tan \theta}{1- \tan^2 \theta} + \tan \theta}{1 - \frac{2 \tan^2 \theta}{1- \tan^2 \theta}} \\ &\equiv \frac{2\tan \theta + \tan \theta -\tan^3 \theta}{1 - \tan^2 \theta - 2 \tan^2 \theta} \\ &\equiv \frac {3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \end{align*} Let \(t = \tan \theta\), then \begin{align*} && \frac{11}{2} &= \frac{3t - t^3}{1-3t^2} \\ \Leftrightarrow && 11 - 33t^2 &= 6t -2t^3 \\ \Leftrightarrow && 0 &= 2t^3-33t^2-6t+11 \\ \Leftrightarrow && 0 &= (2t-1)(t^2-16t-11) \end{align*} Therefore \(\tan \theta = \frac12, \tan \theta = \frac{16 \pm \sqrt{16^2+4 \cdot 1 \cdot 11}}{2} = \frac{16\pm10\sqrt{3}}{2} = 8 \pm 5 \sqrt{3}\). Since \(\displaystyle \frac{\pi}{ 4} \le \theta \le \frac{\pi}{2}\) we must have that \(\tan\) is both positive and \(\geq 1\), therefore \(\tan \theta = 8 + 5 \sqrt{3}\)

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2003 Paper 2 Q2
D: 1600.0 B: 1500.0
trigonometry harmonic form R-alpha method arctan arccos arcsin exact values identity manipulation

Write down a value of \(\theta\,\) in the interval \(\frac{1}{4}\pi< \theta <\frac{1}{2}\pi\) that satisfies the equation \[ 4\cos\theta+ 2\sqrt3\, \sin\theta = 5 \;. \] Hence, or otherwise, show that \[ \pi=3\arccos(5/\sqrt{28}) + 3\arctan(\sqrt3/2)\;. \] Show that \[ \pi=4\arcsin(7\sqrt2/10) - 4\arctan(3/4)\;. \]

Solution: If \(\theta = \frac{\pi}{3}\) then \(\cos \theta = \frac12, \sin \theta = \frac{\sqrt{3}}{2}\) and clearly the equation is satisfied. We can also solve this equation using the harmonic formulae, namely: \begin{align*} && 5 &= 4 \cos \theta + 2\sqrt{3} \sin \theta \\ &&&= \sqrt{4^2+2^2 \cdot 3} \cos \left (\theta -\tan^{-1} \left (\frac{2\sqrt{3}}{4}\right) \right) \\ \Rightarrow && \frac{5}{\sqrt{28}} &= \cos \left ( \frac{\pi}{3} - \tan^{-1} \left (\frac{\sqrt{3}}{2}\right) \right) \\ \Rightarrow && \frac{\pi}{3} &= \arccos\left( \frac{5}{\sqrt{28}}\right) + \arctan \left (\frac{\sqrt{3}}{2}\right) \end{align*} From which the result follows. Similarly, notice that \(3 \cos \theta + 4 \sin \theta = \frac{7}{\sqrt{2}}\) is clearly solved by \(\frac{\pi}{4}\), but also writing it in harmonic form, we have \begin{align*} &&\frac{7}{\sqrt{2}} &= 5 \sin \left (\theta + \tan^{-1} \left ( \frac{3}{4} \right) \right) \\ \Rightarrow && \frac{7\sqrt{2}}{10} &= \sin \left ( \frac{\pi}{4} + \tan^{-1} \left ( \frac{3}{4} \right) \right) \\ \Rightarrow && \frac{\pi}{4} &= \arcsin \left ( \frac{7\sqrt{2}}{10} \right) - \arctan \left ( \frac{3}{4} \right) \end{align*} as required.

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2001 Paper 1 Q4
D: 1500.0 B: 1484.0
trigonometry tan 3θ identity triple angle formula inverse trigonometric functions solving trigonometric equations exact values compound angle formula

Show that \(\displaystyle \tan 3\theta = \frac{3\tan\theta -\tan^3\theta}{1-3\tan^2\theta}\) . Given that \(\theta= \cos^{-1} (2/\sqrt5)\) and \(0<\theta<\pi/2\), show that \(\tan 3\theta =11/2\) Hence, or otherwise, find all solutions of the equations

  1. \(\tan(3\cos^{-1} x) =11/2\) ,
  2. \(\cos ({\frac13}\tan^{-1} y) = 2/\sqrt5\) .

Solution: Let \(\tan \theta = t\) \begin{align*} \tan 3 \theta &\equiv \tan (2 \theta + \theta) \\ &\equiv \frac{\tan 2 \theta +\tan \theta}{1 - \tan 2 \theta \tan \theta} \\ &\equiv \frac{\frac{2t}{1-t^2}+t}{1-\frac{2t^2}{1-t^2}} \\ &\equiv \frac{2t+t-t^3}{1-t^2-2t^2} \\ &\equiv \frac{3t-t^3}{1-3t^3} \\ &\equiv \frac{3\tan \theta - \tan^3 \theta}{1 - 3 \tan^3 \theta} \end{align*} If \(\theta = \cos^{-1} (2/\sqrt{5})\), then \(\sin \theta = 1/\sqrt{5}\) and \(\tan \theta = 1/2\). Hence \begin{align*} \tan 3 \theta &= \frac{3 \cdot \frac12 - \frac18}{1 - \frac34} \\ &= \frac{11}{2} \end{align*}

  1. Since \(\tan 3 y = 11/2\) has the solution \(y = \cos^{-1} (2/\sqrt{5})\) it will also have the solutions \(y = \cos^{-1}(2/\sqrt{5}) + \frac{\pi}{3}\) and \(y = \cos^{-1}(2/\sqrt{5})+\frac{2\pi}{3}\), therefore \begin{align*} && \cos^{-1} x &= \cos^{-1} (2/\sqrt{5})\\ \Rightarrow && x &= 2/\sqrt{5} \\ && \cos^{-1} x &= \cos^{-1} (2/\sqrt{5}) + \frac{\pi}{3}\\ \Rightarrow && x &= \frac{2}{\sqrt{5}} \frac{1}{2} - \frac{1}{\sqrt{5}} \frac{\sqrt{3}}{2} \\ &&&= \frac{2-\sqrt{3}}{2\sqrt{5}} \\ && \cos^{-1} x &= \cos^{-1} (2/\sqrt{5}) + \frac{2\pi}{3}\\ \Rightarrow && x &= \frac{2}{\sqrt{5}} \left (-\frac{1}{2} \right)- \frac{1}{\sqrt{5}} \frac{\sqrt{3}}{2} \\ &&&= \frac{-\sqrt{3}-2}{2\sqrt{5}} \\ \end{align*}
  2. Since \(\cos \frac13 x = \frac{2}{\sqrt{5}}\) has the solution \(x = \tan^{-1} \frac{11}{2}\) it will also have the solutions \(x = \tan^{-1} \frac{11}{2} + 2n \pi\) and \(x = -\tan^{-1} \frac{11}{2} + 2n \pi\). \begin{align*} && \tan^{-1} y &= \tan^{-1} \frac{11}{2} \\ \Rightarrow && y &= \frac{11}{2} \\ && \tan^{-1} y &= \tan^{-1} \frac{11}{2} + 2n \pi \\ \Rightarrow && y &= \frac{\frac{11}{2} + 0}{1-0} \\ &&&= \frac{11}{2} \\ && \tan^{-1} y &= -\tan^{-1} \frac{11}{2} + 2n \pi \\ \Rightarrow && y &= \frac{-\frac{11}{2} + 0}{1-0} \\ &&&= -\frac{11}{2} \\ \end{align*} So our two solutions are \(y = \pm \frac{11}{2}\)

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