2012 H2 Mathematics Paper 1 Question 4

Maclaurin Series

Answers

(ii)

1+θ+32θ2{1 + \theta + \frac{3}{2} \theta^2}

Full solutions

(i)

By Sine Rule,
ACBC=sin⁡34πsin⁡(π−34π−θ)AC1=22sin⁡(14π−θ)AC=22sin⁡14πcos⁡θ−cos⁡14πsin⁡θ=2222cos⁡θ−22sin⁡θ=1cos⁡θ−sin⁡θ  ■\begin{align*} \frac{AC}{BC} &= \frac{\sin \frac{3}{4}\pi}{\sin \left( \pi - \frac{3}{4}\pi - \theta \right)} \\ \frac{AC}{1} &= \frac{\frac{\sqrt{2}}{2}}{\sin \left( \frac{1}{4}\pi - \theta \right)} \\ AC &= \frac{\frac{\sqrt{2}}{2}}{\sin \frac{1}{4}\pi \cos \theta - \cos \frac{1}{4}\pi \sin \theta} \\ &= \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}\cos \theta - \frac{\sqrt{2}}{2} \sin \theta} \\ &= \frac{1}{\cos \theta - \sin \theta} \; \blacksquare \end{align*}

(ii)

AC=1cos⁡θ−sin⁡θ≈1(1−12θ2)−θ=(1−θ−12θ2)−1≈1+(−1)(−θ−12θ2)+(−1)(−2)2(−θ−12θ2)2≈1+θ+12θ2+θ2=1+θ+32θ2  ■\begin{align*} AC &= \frac{1}{\cos \theta - \sin \theta} \\ &\approx \frac{1}{\left(1 - \frac{1}{2} \theta^2\right) - \theta} \\ &= \left( 1 - \theta - \frac{1}{2} \theta^2 \right)^{-1} \\ &\approx 1 + (-1)\left( - \theta - \frac{1}{2} \theta^2 \right) + \frac{(-1)(-2)}{2} \left( - \theta - \frac{1}{2} \theta^2 \right)^2 \\ &\approx 1 + \theta + \frac{1}{2}\theta^2 + \theta^2 \\ &= 1 + \theta + \frac{3}{2} \theta^2 \; \blacksquare \end{align*}