(a) Simplify : \(\frac{1}{1 - \cos \theta} + \frac{1}{1 + \cos \theta}\) and leave your answer in terms of \(\sin \theta\). (b) Find the equation of the lin...
Question 1 Report
(a) Simplify : \(\frac{1}{1 - \cos \theta} + \frac{1}{1 + \cos \theta}\) and leave your answer in terms of \(\sin \theta\).
(b) Find the equation of the line joining the stationary points of \(y = x^{2} (x - 3)\) and the distance between them.
Answer Details
(a) Add the two fractions over a common denominator:
\[\frac{1}{1-\cos\theta}+\frac{1}{1+\cos\theta}=\frac{(1+\cos\theta)+(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}=\frac{2}{1-\cos^{2}\theta}\]
Since \(1-\cos^{2}\theta=\sin^{2}\theta\):
\[=\frac{2}{\sin^{2}\theta}\]
(b) \(y=x^{2}(x-3)=x^{3}-3x^{2}\). Then \(\dfrac{dy}{dx}=3x^{2}-6x=3x(x-2)\).
Stationary points where \(\dfrac{dy}{dx}=0\): \(x=0\) or \(x=2\).
At \(x=0,\ y=0\Rightarrow(0,0)\). At \(x=2,\ y=8-12=-4\Rightarrow(2,-4)\).
Line joining them: slope \(=\dfrac{-4-0}{2-0}=-2\), through \((0,0)\):
\[y=-2x\quad\text{or}\quad 2x+y=0\]
Distance:
\[\sqrt{(2-0)^{2}+(-4-0)^{2}}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}\]
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