Question 11.4.23

The triple angle formula states that \[\cos(3\theta) = 4\cos^3\theta - 3\cos\theta.\]

1. We’re going to solve the cubic equation $x^3 = 3px + 2q$ for $x$ using the triple angle formula. If $x=2r\cos\theta$, find values of $r$ and $\theta$ that will solve the cubic equation. (In Question 7.5.34 we were able to solve this equation when $q^2-p^3 \ge 0$.) Show answer

   \begin{align*} x^3 &= 8r^3\cos^3\theta\\ &= 2r^3\times4\cos^3\theta\\ &= 2r^3\left(3\cos\theta + \cos(3\theta)\right)\\ &= 2r^33\cos\theta + 2r^3\cos(3\theta)\\ &= 3 r^2 2r\cos\theta + 2 r^3\cos(3\theta)\\ &= 3 r^2 x + 2r^3\cos(3\theta)\\ &= 3px + 2q. \end{align*} So $p=r^2$ and $q=r^3\cos(3\theta)$. Hence $r=\sqrt p$ and $\theta = \arccos(q/r^3)/3$.

   

2. What values of $p$ and $q$ will your solution work for? Show hint

   What values can $r^2$ and $\cos3\theta$ take?

   Show answer

   $r^2$ can’t be negative, so $p\ge0$.

   $\cos3\theta = q/\sqrt p^3$, and the magnitude of $\cos 3\theta$ must be less than $1$, so: \begin{align*} \abs{\frac{q}{\sqrt p^3}} &\le 1\\ \frac{q^2}{p^3} &\le 1\\ q^2 &\le p^3. \end{align*}

   

3. Using your solution, find all values of $x$ that satisfy the equation $x^3 = 6x + 4$. Show hint

   Cubic equations can have three solutions.

   Show answer

   In this case, $3p=6$ and $2q=4$, so $p=2$ and $q=2$. \begin{align*} r &= \sqrt p\\ &= \sqrt2\\ \cos(3\theta) &= \frac{q}{r^3}\\ &= \frac{2}{(\sqrt2)^3}\\ &= \frac{1}{\sqrt2} \end{align*} Now here’s where we need to be careful about finding all the solutions. The general solution gives: \begin{align*} 3\theta &= 2n\pi \pm \frac{\pi}{4}\\ \theta &= \frac{2n\pi}{3}\pm \frac{\pi}{12}\\ &= \frac{(8n \pm 1)\pi}{12} \end{align*} You can see these angles depicted in the following diagram, with $\theta_n = (8n+1)\pi/12$ and $\hat\theta_n = (8n-1)\pi/12$ shown in red: $\theta_{n+3}$ differs from $\theta_n$ by $2\pi$ so they’re coterminal, which is why the diagram only shows the angles for $n=0,1,2$ – the others are repeats. The red ones are all mirror images of the black ones reflected about the $x$-axis because we always get positive and negative pairs of angles when we use arccos (e.g. $\hat \theta_2$ is coterminal with $\hat\theta_{-1} = -9\pi/12$ which is a mirror image of $\theta_1 = 9\pi/12$).

   There are a couple approaches you might use here to decide which angles give unique solutions. Since $x = 2r\cos\theta$, those angles below the $x$-axis will give the same solutions as the ones above the $x$-axis ($\cos\theta_0 = \cos\hat\theta_0$, $\cos\theta_1 = \cos\hat\theta_2$, and $\cos\theta_2 = \cos\hat\theta_1$), so we might decide that we’re only interested in values of $\theta$ between $0$ and $\pi$ ($\theta_0$, $\hat\theta_1$ and $\theta_1$ on the diagram). So we need to find all solutions in the range $0\le 3\theta \le 3\pi$. This gives: \begin{align*} 3\theta &= \frac{\pi}{4},\frac{7\pi}{4},\frac{9\pi}{4}\\ \theta &= \frac{\pi}{12},\frac{7\pi}{12},\frac{3\pi}{4}\\ x &= 2\sqrt2\cos\frac{\pi}{12},2\sqrt2\cos\frac{7\pi}{12},2\sqrt2\cos\frac{3\pi}{4}. \end{align*} That last solution can be simplified to $x=-2$.

   Or, you might decide that the red angles give the same solutions as the black ones, so we’ll ignore the $3\theta = 2n\pi - \pi/4$ case. And we only need to go up to $n=2$ because $\theta_3 = \theta_0 + 2\pi$. So this gives: \begin{align*} 3\theta &= \frac{\pi}{4},\frac{\pi}{4}+2\pi,\frac{\pi}{4} + 4\pi\\ \theta &= \frac{\pi}{12},\frac{3\pi}{4},\frac{17\pi}{12}\\ x &= 2\sqrt2\cos\frac{\pi}{12}, 2\sqrt2\cos\frac{3\pi}{4},2\sqrt2\cos\frac{17\pi}{12}. \end{align*} Again, $2\sqrt2\cos\frac{3\pi}{4}$ simplifies to $-2$.

   In any case you should have three solutions: $-2$, $2\sqrt2\cos(\pi/12)$ and either $2\sqrt2\cos(7\pi/12)$ or $2\sqrt2\cos(17\pi/12)$ depending on how you chose the angles (but it doesn’t matter because they’re the same number). For those of you with calculators, those solutions to two decimal places are \[x=-2,-0.73,2.73.\]