1 problem found
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