SPM Additional Mathematics 2019, Paper 2 (Question 10)

Question 10:

$\begin{array}{l} (a) (i) Prove t a n \frac{A}{2} = \frac{1 - \cos A}{\sin A} . \\ (ii) {Hence, without using calculator, find the value of tan15}^{o} . \\ State your answer in the form p - \sqrt{q}, where p and q are \\ constants . \\ (b) (i) Sketch the graph of y = - \frac{3}{2} \sin A for 0 \leq A \leq 2 π . \\ (ii) Hence, using the same axes, sketch a suitable straight \\ line to find the number of solutions for the equation \\ (\cot \frac{A}{2}) (1 - \cos A) = - \frac{A}{2 π} for 0 \leq A \leq 2 π . \\ State the number solutions . \end{array}$

Solution:
(a)(i)

$\begin{array}{l} Right hand side, \\ \frac{1 - \cos A}{\sin A} = \frac{2 \sin^{2} \frac{A}{2}}{2 \sin \frac{A}{2} \cos \frac{A}{2}} \\ = \frac{\sin \frac{A}{2}}{\cos \frac{A}{2}} \\ = \tan \frac{A}{2} \\ = (Left hand side) \end{array}$

(a)(ii)

$\begin{array}{l} \tan (\frac{30}{2}) = \frac{1 - \cos 30^{o}}{\sin 30^{o}} \\ t a n 15^{o} = \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}} \\ = \frac{\frac{2 - \sqrt{3}}{2}}{\frac{1}{2}} \\ = 2 - \sqrt{3} \end{array}$

(b)(i)

(b)(ii)

$\begin{array}{l} (\cot \frac{A}{2}) (1 - \cos A) = - \frac{A}{2 π} \\ \frac{1}{\tan \frac{A}{2}} (1 - \cos A) = - \frac{A}{2 π} \\ (\frac{\sin A}{1 - \cos A}) (1 - \cos A) = - \frac{A}{2 π} \\ \sin A = - \frac{A}{2 π} \\ - \frac{3}{2} \sin A = - \frac{A}{2 π} \times (- \frac{3}{2}) \\ y = \frac{3 A}{4 π} \end{array}$

Number of solutions = 3

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