# 2005_Question2

(i)

Note that:

1^3 - (1 + \cos \theta + \sin \theta) + (\cos \theta \sin \theta + \cos \theta + \sin \theta) - \sin \theta \cos \theta = 0

So the cubic has a root at x = 1, so we can write it in the form:

(x - 1)(x^2 + Ax + B)

since it has leading coefficient 1. Expanding:

(x - 1)(x^2 + Ax + B) = x^3 + Ax^2 + B x - x^2 - Ax - B

We want -B = -\sin \theta \cos \theta, so B = \sin \theta \cos \theta. Then we have B - A = \cos \theta \sin \theta + \cos \theta + \sin \theta, so A = -(\cos \theta + \sin \theta). We can see that agrees with the x^2 coefficient, so the cubic (1) can be written as:

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

We can factorise the quadratic as:

(x - \cos \theta)(x - \sin \theta)

(since the roots sum to \sin \theta + \cos \theta and multiply to \sin \theta \cos \theta, and the leading coefficient is 1)

So we can write the cubic as:

(x - 1)(x - \cos \theta)(x - \sin \theta)

where we can straightforwardly read off the roots x = 1, x = \cos \theta and x = \sin \theta.

(ii)

The roots are 1, \displaystyle \sin \frac \pi 3 = \frac {\sqrt 3} 2, \displaystyle \cos \frac \pi 3 = \frac 1 2.

(iii) We can either have:

1 = \cos \theta

giving \theta = 0, or:

1 = \sin \theta

giving \theta = \dfrac \pi 2, or:

\sin \theta = \cos \theta

which is equivalent to \tan \theta = 1 (since \sin and \cos are never simultaneously zero), giving \theta = \dfrac \pi 4 or \theta = \dfrac {5 \pi} 4.

(iv)

We have that |1 - \cos \theta| \le 2 and |1 - \sin \theta| \le 2. (since -1 \le \cos \theta \le 1, giving -2 \le \cos \theta - 1 \le 0 and similarly for \sin) Note that these bounds are attained with \sin \theta = -1 and \cos \theta = -1. We have this if x = \dfrac {3 \pi} 2 or x = \pi.

Note that:

\displaystyle \cos \theta - \sin \theta = \sqrt 2 \left(\frac {\sqrt 2} 2 \cos \theta - \frac {\sqrt 2} 2 \sin \theta\right) = \sqrt 2 \cos \left(\theta + \frac \pi 4\right)

so:

\displaystyle |\cos \theta - \sin \theta| \le \sqrt 2 < 2 - so the greatest possible difference is 2, occuring at \theta = \pi and \theta = \dfrac {3 \pi} 2.

To show that the cubics are the same for these values of \theta, it suffices to show that \cos \theta + \sin \theta and \cos \theta \sin \theta are the same for each of these values of \theta. (since we have the factorisation (x - 1)(x^2 - (\cos \theta + \sin \theta)x + \cos \theta \sin \theta).

In the case \theta = \pi we have:

\cos \theta + \sin \theta = -1 + 0 = -1

and in the case \theta = \dfrac {3 \pi} 2 we have:

\cos \theta + \sin \theta = 0 - 1 = -1

We hence have \sin \theta \cos \theta = 0 in both cases. So the cubic is the same in both cases as desired.