- Show, with working, that \[\begin{equation} x^3 - (1 + \cos\theta + \sin\theta)x^2 + (\cos\theta\sin\theta + \cos\theta + \sin\theta)x - \sin\theta\cos\theta, \label{eq:1repeat} \end{equation}\] equals \[(x - 1)(x^2 - (\cos\theta + \sin\theta)x + \cos\theta\sin\theta).\]
Deduce that the cubic in \(\eqref{eq:1repeat}\) has roots \[1, \qquad \cos\theta, \qquad \sin\theta.\]
If we factorise the quadratic above further we get \[x^2 - (c+s)x + cs = (x-c)(x-s),\] and so the three roots of the cubic are \(1\), \(c\) and \(s\).
- Give the roots when \(\theta = \dfrac{\pi}{3}\).
In this case, the three roots are \(1\), \(c = \cos \dfrac{\pi}{3} = \dfrac{1}{2}\) and \(s = \sin \dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}\).
- Find all values of \(\theta\) in the range \(0 \leq \theta < 2\pi\) such that two of the three roots are equal.
There are three possible cases:
- \(s = 1\), which only occurs in the given range when \(\theta = \pi/2\);
- \(c = 1\), in which case \(\theta = 0\);
- \(s = c\), that is, \(\tan\theta = 1\) and \(\theta = \pi/4\) or \(5\pi/4\).
- What is the greatest possible difference between two of the roots, and for what values of \(\theta\) in the range \(0 \leq \theta < 2\pi\) does this greatest difference occur?
There are three cases.
- we take the difference between \(1\) and \(s\), which is largest when \(s = -1\), in which case the difference is \(2\).
- we take the difference between \(1\) and \(c\), which is largest when \(c = -1\), in which case the difference is \(2\).
- we take the difference between \(c\) and \(s\). Here it is impossible for one to be \(1\) and the other to be \(-1\) at the same time, so their difference must be less than \(2\).
More precisely, we could say that for case (3), the difference between \(c\) and \(s\) is largest when \(\big\vert c - s\big\vert\) is largest, which is when \(\sqrt{2}\big\vert\cos \big(x + \dfrac{\pi}{4}\big)\big\vert\) is largest.
The largest possible value here is \(\sqrt{2} < 1\), so we can rule this out.
Cases (1) and (2) occur at \(\theta = 3\pi/2\) and \(\theta = \pi\) respectively.
Show that for each such \(\theta\) the cubic \(\eqref{eq:1repeat}\) is the same.
When \(c = -1\) then \(s = 0\), and when \(s = -1\) then \(c = 0\). Since the cubic is symmetric in \(s\) and \(c\), the cubic will be the same in each case.
We can calculate the polynomial explicitly in each case as a check to get
\[(x-1)(x-0)(x+1) = x^3 - x.\]
