Problems

Filters
Clear Filters

9 problems found

2023 Paper 2 Q2
D: 1500.0 B: 1500.0
trigonometric identities double angle formula triple angle formula simultaneous equations tan substitution polynomial degree chebyshev polynomials

  1. The real numbers \(x\), \(y\) and \(z\) satisfy the equations \[y = \frac{2x}{1-x^2}\,,\qquad z = \frac{2y}{1-y^2}\,,\qquad x = \frac{2z}{1-z^2}\,.\] Let \(x = \tan\alpha\). Deduce that \(y = \tan 2\alpha\) and show that \(\tan\alpha = \tan 8\alpha\). Find all solutions of the equations, giving each value of \(x\), \(y\) and \(z\) in the form \(\tan\theta\) where \(-\frac{1}{2}\pi < \theta < \frac{1}{2}\pi\).
  2. Determine the number of real solutions of the simultaneous equations \[y = \frac{3x - x^3}{1-3x^2}\,,\qquad z = \frac{3y - y^3}{1-3y^2}\,,\qquad x = \frac{3z - z^3}{1-3z^2}\,.\]
  3. Consider the simultaneous equations \[y = 2x^2 - 1\,,\qquad z = 2y^2 - 1\,,\qquad x = 2z^2 - 1\,.\]
    1. Determine the number of real solutions of these simultaneous equations with \(|x| \leqslant 1\), \(|y| \leqslant 1\), \(|z| \leqslant 1\).
    2. By finding the degree of a single polynomial equation which is satisfied by \(x\), show that all solutions of these simultaneous equations have \(|x| \leqslant 1\), \(|y| \leqslant 1\), \(|z| \leqslant 1\).

View
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*}

View
2011 Paper 1 Q3
D: 1500.0 B: 1500.0
trigonometry trig identity product-to-sum differentiation cotangent cosecant triple angle formula proof

Prove the identity \[ 4\sin\theta \sin(\tfrac13\pi-\theta) \sin (\tfrac13\pi+\theta)= \sin 3\theta\, . \tag{\(*\)}\]

  1. By differentiating \((*)\), or otherwise, show that \[ \cot \tfrac19\pi - \cot \tfrac29\pi + \cot \tfrac49\pi = \sqrt3\,. \]
  2. By setting \(\theta = \frac16\pi-\phi\) in \((*)\), or otherwise, obtain a similar identity for \(\cos3\theta\) and deduce that \[ \cot \theta \cot (\tfrac13\pi-\theta) \cot (\tfrac13\pi+\theta) =\cot3\theta\,. \] Show that \[ \cosec \tfrac19\pi -\cosec \tfrac59\pi +\cosec \tfrac79\pi = 2\sqrt3\,. \]

Solution: \begin{align*} && LHS &= 4\sin\theta \sin(\tfrac13\pi-\theta) \sin (\tfrac13\pi+\theta) \\ &&&= 4 \sin \theta \left (\tfrac{\sqrt{3}}{2}\cos \theta - \tfrac12 \sin \theta \right)\left (\tfrac{\sqrt{3}}{2}\cos \theta + \tfrac12 \sin \theta \right) \\ &&&= 4 \sin \theta \left (\tfrac{3}{4}\cos^2 \theta - \tfrac14 \sin^2 \theta \right) \\ &&&= 3\sin \theta - 4\sin^3 \theta \\ &&&= \cos 3 \theta = RHS \end{align*}

  1. \(\,\) \begin{align*} && 3 \cos 3 \theta &= \sin 3 \theta \left (\cot \theta - \cot (\tfrac13\pi - \theta) + \cot (\tfrac13\pi + \theta) \right) \\ \Rightarrow && 3 \cot 3\theta &= \cot \theta - \cot (\tfrac13\pi - \theta) + \cot (\tfrac13\pi + \theta) \\ \theta = \tfrac{\pi}{9}: && 3\cot \frac{\pi}{3} &= \cot \tfrac{\pi}{9} - \cot \tfrac{2}{9}\pi + \cot \tfrac49 \pi \\ \Rightarrow && \sqrt{3} &= \cot \tfrac{\pi}{9} - \cot \tfrac{2}{9}\pi + \cot \tfrac49 \pi \end{align*}
  2. \(\,\) \begin{align*} \theta = \tfrac16\pi - \phi && \sin(\tfrac12\pi - 3\phi) &= 4\sin(\tfrac16\pi - \phi)\sin(\phi+\tfrac16\pi)\sin(\tfrac12\pi - \phi) \\ \Rightarrow && \cos 3\phi &= 4\cos(\phi - \tfrac13\pi)\cos(\tfrac13\pi - \phi)\cos\phi \\ \Rightarrow && \cot 3\theta &= \cot \theta\cot(\phi - \tfrac13\pi)\cot(\tfrac13\pi - \phi) \tag{dividing by (\(*\))} \\ \\ \frac{\d}{\d \theta}:&& -\csc^2 3\phi &= \cot3\phi \left (-\csc^2 \phi\tan \phi+\csc^2 (\tfrac13\pi - \phi) \tan (\tfrac13\pi - \phi) -\csc^2(\phi - \tfrac13\pi)\tan (\phi - \tfrac13\pi) \right) \\ \Rightarrow && \csc^23\phi\tan3\phi & = 2( \csc2\phi- \csc(\tfrac{2}{3}\pi - 2\phi)+\csc(\phi - \tfrac23\pi)) \\ \phi = \frac{1}{18}\pi: && 4\sqrt{3} &= 2(\csc \tfrac{1}{9}\pi - \csc \tfrac59\pi + \csc \tfrac79 \pi) \\ \end{align*} and the result follows.

View
2010 Paper 2 Q2
D: 1600.0 B: 1506.8
integration trigonometric triple angle formula cos 3x sin 3x polynomial equation cubic definite integral proof

Prove that \[ \cos 3x = 4 \cos^3 x - 3 \cos x \,. \] Find and prove a similar result for \(\sin 3x\) in terms of \(\sin x\).

  1. Let \[ {\rm I}(\alpha) = \int_0^\alpha \big(7\sin x - 8 \sin^3 x\big) \d x\,. \] Show that \[ {\rm I}(\alpha) = -\tfrac 8 3 c^3 + c +\tfrac5 3\,, \] where \(c = \cos \alpha\). Write down one value of \(c\) for which \({\rm I}(\alpha) =0\).
  2. Useless Eustace believes that \[ \int \sin^n x \, \d x =\dfrac {\sin^{n+1}x}{n+1}\, \] for \(n=1, \ 2, \ 3, \ldots\, \). Show that Eustace would obtain the correct value of \({\rm I}(\beta)\,\), where \(\cos \beta= -\frac16\). Find all values of \(\alpha\) for which he would obtain the correct value of \({\rm I}(\alpha)\).

Solution: \begin{align*} \cos 3x &\equiv \cos (2x + x) \\ &\equiv \cos 2x \cos x - \sin 2x \sin x \\ &\equiv (2\cos^2 x - 1) \cos x - 2 \sin x \cos x \sin x \\ &\equiv 2 \cos^3 x - \cos x - 2\cos x (\sin^2 x) \\ &\equiv 2 \cos^3 x - \cos x - 2\cos x (1- \cos^2 x) \\ &\equiv 4\cos^3 x - 3\cos x \end{align*} Similarly, \begin{align*} \sin 3x &\equiv \sin (2x + x) \\ &\equiv \sin 2x \cos x + \cos 2x \sin x \\ &\equiv 2 \sin x \cos x \cos x + (1-2\sin^2 x) \sin x \\ &\equiv 2 \sin x (1-\sin^2 x) + \sin x - 2 \sin^3 x \\ &\equiv 3 \sin x -4 \sin ^3 x \end{align*}

  1. \begin{align*} I(\alpha) &= \int_0^{\alpha} (7 \sin x - 8 \sin^3 x) \d x \\ &= \int_0^{\alpha} (7 \sin x - (6\sin x-2 \sin 3x) ) \d x \\ &= \int_0^{\alpha} (\sin x +2 \sin 3x ) \d x \\ &= -\cos \alpha - \frac23 \cos 3\alpha +1+\frac23 \\ &= -c - \frac23 (4c^3-3c) + \frac53 \\ &= -\frac83 c^3 +c + \frac53 \end{align*} as required. When \(c = -1\) this value is \(0\). Eustace will obtain the value \(\frac{7}{2} \sin^2 \beta - 2 \sin^4 \beta = \frac72 (1-\cos^2 \beta) - 2(1-\cos^2 \beta)^2 = \frac32 + \frac12\cos^2 \beta -2\cos^4 \beta\) So if \(\cos \beta = -\frac16\) he will obtain \(\frac32 + \frac{1}{2\cdot36} - \frac{2}{6^4}\) and he should obtain \(\frac{8}{3} \frac{1}{6^3} - \frac{1}{6} + \frac{5}{3}\) which are equal. We want to find all roots of: \begin{align*} && \frac32 + \frac12 c^2 - 2c^4 &= -\frac83 c^3+ c + \frac53 \\ \Rightarrow && 0 &=2c^4-\frac83c^3-\frac12 c^2+c +\frac{1}{6} \\ &&&= 12c^4-16c^3-3c^2+6c+1\\ &&&= (6c+1)(2c^3-3c^2+1) \\ &&&= (6c+1)(2c+1)(c-1)^2 \end{align*} Therefore \(\cos \alpha = - \frac16, -\frac12, 1\) will give the correct answers.

View
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 \(\tan \theta = 8 + 5 \sqrt{3}\)

View
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 \(y = \pm \frac{11}{2}\)

View
1999 Paper 1 Q7
D: 1500.0 B: 1516.0
second order differential equation trigonometric inverse trigonometric chain rule polynomial solution verification triple angle formula sin 3θ identity

Show that \(\sin(k\sin^{-1} x)\), where \(k\) is a constant, satisfies the differential equation $$(1-x^{2})\frac {\d^2 y}{\d x^2} -x\frac{\d y}{\d x} +k^{2}y=0. \tag{*}$$ In the particular case when \(k=3\), find the solution of equation \((*)\) of the form \[ y=Ax^{3}+Bx^{2}+Cx+D, \] that satisfies \(y=0\) and \(\displaystyle \frac{\d y}{\d x}=3\) at \(x=0\). Use this result to express \(\sin 3\theta\) in terms of powers of \(\sin\theta\).

Solution: \begin{align*} && y &= \sin(k \sin^{-1} x ) \\ &&y' &= \cos (k \sin^{-1} x) \cdot k \frac{1}{\sqrt{1-x^2}} \\ && y'' &= -\sin (k \sin^{-1} x) \cdot k^2 \frac{1}{(1-x^2)} - \cos(k \sin^{-1} x) \cdot k \frac{x}{(1-x^2)\sqrt{1-x^2}} \\ && (1-x^2)y'' &= -k^2y -xy' \\ \Rightarrow && 0 &= (1-x^2)y''+xy' + k^2y \end{align*} \begin{align*} && y &= Ax^3 + Bx^2 + Cx + D \\ && y' &= 3Ax^2 + 2Bx + C \\ && y'' &= 6Ax+2B \\ && 0 &= (1-x^2)(6Ax+2B) - x( 3Ax^2 + 2Bx + C) + 9(Ax^3 + Bx^2 + Cx + D ) \\ &&&= x^3(-6A-3A+9A) + x^2(-2B-2B+9B) + x(6A-C+9C) + (2B +9D) \\ \Rightarrow && B &= 0 \\ \Rightarrow && D &= 0 \\ \Rightarrow && C &= -\frac34 A \\ \\ x = 0, y = 0, y' = 0: && y &= 3x-4x^3 \\ \end{align*} And so \(\sin 3 x = 3 \sin x - 4\sin^3 x\)

View
1997 Paper 2 Q6
D: 1600.0 B: 1500.0
trigonometry tan identity double angle formula triple angle formula trigonometric equations cot quadratic in tan

Show that, if \(\,\tan^2\phi=2\tan\phi+1\), then \(\tan2\phi=-1\). Find all solutions of the equation $$\tan\theta=2+\tan3\theta$$ which satisfy \(0<\theta< 2\pi\), expressing your answers as rational multiples of \(\pi\). Find all solutions of the equation the equation $$\cot\theta=2+\cot3\theta$$ which satisfy $$-\frac{3\pi}{2}<\theta<\frac{\pi}{2}.$$

View
1989 Paper 2 Q1
D: 1600.0 B: 1516.0
trigonometry cos nθ triple angle formula cubic equation substitution solving cubics via trig Chebyshev polynomial

Prove that \(\cos3\theta=4\cos^{3}\theta-3\cos\theta\). Show how the cubic equation \[ 24x^{3}-72x^{2}+66x-19=0\tag{*} \] can be reduced to the form \[ 4z^{3}-3z=k \] by means of the substitution \(y=x+a\) and \(z=by\), for suitable values of the constants \(a\) and \(b\). Hence find the three roots of the equation \((*)\), to three significant figures. Show, by means of a counterexample, or otherwise, that not all cubic equations of the form \[ x^{3}+\alpha x^{2}+\beta x+\gamma=0 \] can be solved by this method.

Solution: \begin{align*} \cos 3\theta &= \cos 2\theta\cos\theta - \sin 2\theta \sin \theta \\ &= (2\cos^2\theta-1)\cos \theta - 2\cos \theta \sin^2 \theta \\ &= 2\cos^3\theta-\cos \theta - 2\cos \theta(1- \cos^2 \theta) \\ &= 4\cos^3 \theta - 3\cos \theta \end{align*} \begin{align*} 0 &= 24x^{3}-72x^{2}+66x-19 \\ &= 24(y+1)^3-72(y+1)^2+66(y+1)-19 \\ &= 24(y^3+3y^2+3y+1)-72(y^2+2y+1)+66(y+1)-19\\ &= 24y^3+(72-144+66)y+(24-72+66-19) \\ &= 24y^3-6y-1 \\ &= 24b'^3z^3 - 6b'z - 1 \\ &= \frac{2}{\sqrt{3}}(4 z^3 -3z) - 1 \\ \end{align*} Therefore if \(b = \sqrt{3}, a = 1\), we have: \(4z^3 - 3z = \frac{\sqrt{3}}{2}\) So if \(z = \cos \theta \Rightarrow \cos 3\theta = \frac{\sqrt{3}}2 \Rightarrow 3 \theta = \frac{\pi}{6}, \frac{11\pi}{6}, \frac{13\pi}{6} \Rightarrow \theta = \frac{\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}\). Since \(\frac{\cos x}{\sqrt{3}} < 1\) we only need to approximate the first part to 2 significant figures. Therefore: \begin{align*} \sqrt{3} &\approx 1 + \frac{1}{1 + \frac{1}{2+\frac11}} = \frac{7}{4} = 1.75\\ \cos \tfrac{\pi}{18} &\approx \cos \frac{1}{6} \approx 1 - \frac{1}{2} \frac{1}{6^2} = \frac{71}{72} \approx 1 - \frac{1}{70} = 1 - 0.014 = 0.986 \\ \frac{\cos \tfrac{\pi}{18}}{\sqrt{3}} & \approx \frac{.986}{1.75} = 0.57 \\ \end{align*} Final answers: \(1.57, 0.803, 0.629\). We wouldn't be able to solve \(x^3 + 1= 0\) using this method, as we would have 2 non-real roots

View