All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$\mathrm{sin}(r+\frac{1}{2})\theta - \mathrm{sin}(r-\frac{1}{2})\theta$

$= \mathrm{sin}r\theta \mathrm{cos}\frac{\theta}{2} + \mathrm{cos}r\theta \mathrm{sin}\frac{\theta}{2} -[\mathrm{sin}r\theta \mathrm{cos}\frac{\theta}{2} - \mathrm{cos}r\theta \mathrm{sin}\frac{\theta}{2}]$

$= 2\mathrm{cos}r\theta \mathrm{sin}\frac{\theta}{2}$

(ii)

$\sum_{r=1}^{n} \mathrm{cos}r\theta$

$= \sum_{r=1}^{n} \frac{\mathrm{sin}(r+\frac{1}{2})\theta - \mathrm{sin}(r-\frac{1}{2})\theta}{2\mathrm{sin}\frac{\theta}{2}}$

$= \frac{1}{2\mathrm{sin}\frac{\theta}{2}} \sum_{r=1}^{n} \mathrm{sin}(r+\frac{1}{2})\theta - \mathrm{sin}(r-\frac{1}{2})\theta$

$= \frac{1}{2\mathrm{sin}\frac{\theta}{2}} [ \mathrm{sin}(\frac{3\theta}{2}) - \mathrm{sin}(\frac{\theta}{2})]$

$+ \mathrm{sin}(\frac{5\theta}{2}) - \mathrm{sin}(\frac{3\theta}{2})$

$+ \ldots$

$+ \mathrm{sin}(n-\frac{1}{2})\theta - \mathrm{sin}(n-\frac{3}{2})\theta$

$+ \mathrm{sin}(n+\frac{1}{2})\theta - \mathrm{sin}(n-\frac{1}{2})\theta$

$= \frac{1}{2\mathrm{sin}\frac{\theta}{2}} (\mathrm{sin}(n+\frac{1}{2})\theta - \mathrm{sin}(\frac{\theta}{2}))$

(iii)

Let P(n) be the statement $\sum_{r=1}^n \mathrm{sin}r\theta = \frac{\mathrm{cos}\frac{\theta}{2} - \mathrm{cos}(n+\frac{1}{2})\theta}{2\mathrm{sin}\frac{\theta}{2}}, n \in \mathbb{Z}^+$

When $n = 1$,

$\mathrm{LHS}~=~ \mathrm{sin}\theta$

$\mathrm{RHS}~=~ \frac{\mathrm{cos}\frac{\theta}{2} - \mathrm{cos}(1+\frac{1}{2})\theta}{2\mathrm{sin}\frac{\theta}{2}} = sin\theta$

Assume P(k) is true for some positive integers k,

$\sum_{r=1}^k \mathrm{sin}r\theta = \frac{\mathrm{cos}\frac{\theta}{2} - \mathrm{cos}(k+\frac{1}{2})\theta}{2\mathrm{sin}\frac{\theta}{2}}$

To prove P(k+1) is true,

$\sum_{r=1}^{k+1} \mathrm{sin}r\theta = \frac{\mathrm{cos}\frac{\theta}{2} - \mathrm{cos}(k+\frac{3}{2})\theta}{2\mathrm{sin}\frac{\theta}{2}}$

$\mathrm{LHS}~=~ \sum_{r=1}^{k+1} \mathrm{sin}r\theta$

$= \sum_{r=1}^{k} \mathrm{sin}r\theta + \mathrm{sin}(k+1)\theta$

$= \frac{\mathrm{cos}\frac{\theta}{2} - \mathrm{cos}(k+\frac{1}{2})\theta}{2\mathrm{sin}\frac{\theta}{2}} + \mathrm{sin}(k+1)\theta$

$= \frac{\mathrm{cos}\frac{\theta}{2} - \mathrm{cos}(k+\frac{1}{2})\theta + 2\mathrm{sin}(k+1)\theta \mathrm{sin}\frac{\theta}{2}}{2\mathrm{sin}\frac{\theta}{2}}$

$= \frac{\mathrm{cos}\frac{\theta}{2} - \mathrm{cos}(k+\frac{1}{2})\theta - \mathrm{cos}(k+\frac{3}{2})\theta + \mathrm{cos}(k+\frac{1}{2})\theta}{2\mathrm{sin}\frac{\theta}{2}}$

$= \frac{\mathrm{cos}\frac{\theta}{2} - \mathrm{cos}(k+\frac{3}{2})\theta}{2\mathrm{sin}\frac{\theta}{2}}$

$\therefore$, P(k+1) is true.

Since P(1) is true, and P(k) is true $\Rightarrow$ P(k+1) is true, by mathematical induction, P(n) is true for all positive integers n.

### KS Comments:

This question was a difficult question and stunned many students that year. Partly due to the lack of exposure to the use of trigonometry in MI. Students should know that the formulas can be found in MF15. As for (ii), students must take note of the word “Hence” and think of using method of differences. The MI requires students to know how to find the product to sum and sum to product formulas using the MF15, which many students do not realise how to use.

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