Can we show this polynomial has roots $1$, $\cos\theta$ and $\sin\theta$?

Can we show this polynomial has roots $1$, $\cos\theta$ and $\sin\theta$?

Question

1. Show, with working, that

$$\begin{equation} x^3 - (1 + \cos\theta + \sin\theta)x^2 + (\cos\theta\sin\theta + \cos\theta + \sin\theta)x - \sin\theta\cos\theta, \qquad \end{equation}$$

equals $(x - 1)(x^2 - (\cos\theta + \sin\theta)x + \cos\theta\sin\theta).$ Deduce that the cubic in $(1)$ has roots $1, \qquad \cos\theta, \qquad \sin\theta.$

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1. Give the roots when $\theta = \dfrac{\pi}{3}$.

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1. Find all values of $\theta$ in the range $0 \leq \theta < 2\pi$ such that two of the three roots are equal.

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1. 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? Show that for each such $\theta$ the cubic $(1)$ is the same.

Solution

1. Show, with working, that

$$\begin{equation} x^3 - (1 + \cos\theta + \sin\theta)x^2 + (\cos\theta\sin\theta + \cos\theta + \sin\theta)x - \sin\theta\cos\theta, \end{equation}$$

equals $(x - 1)(x^2 - (\cos\theta + \sin\theta)x + \cos\theta\sin\theta).$

The simplest way to show this is to start with the factorised expression and multiply it out. If we write

$c$

for

$\cos\theta$

and

$s$

for

$\sin\theta$

we have

$$\begin{align*} (x - 1)(x^2 - (c + s)x + cs) &= [x^3 - (c+s)x^2 + csx] - [x^2 -(c+s)x + cs]\\ &= x^3 - (1 + c + s)x^2 + (cs + c + s)x - cs. \end{align*}$$

Deduce that the cubic in $(1repeat)$ 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 $s$.

In this case, the three roots are $1$, $c = \cos \dfrac{\pi}{3} = \dfrac{1}{2}$ and $s = \sin \dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$.

There are three possible cases:

1. $s = 1$, which only occurs in the given range when $\theta = \pi/2$;

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1. $c = 1$, in which case $\theta = 0$;

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1. $s = c$, that is, $\tan\theta = 1$ and $\theta = \pi/4$ or $5\pi/4$.

There are three cases.

1. we take the difference between $1$ and $s$, which is largest when $s = -1$, in which case the difference is $2$.

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1. we take the difference between $1$ and $c$, which is largest when $c = -1$, in which case the difference is $2$.

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1. 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 $(1repeat)$ 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

$$(x-1)(x-0)(x+1) = x^3 - x.$$