2011 H2 Mathematics Paper 1 Question 6

Sigma Notation

Answers

(ii)

sin⁡(n+12)θ−sin⁡12θ2sin⁡12θ{\frac{\sin {\left(n+\frac{1}{2}\right)\theta}-\sin { \frac{1}{2}\theta}}{2 \sin { \frac{1}{2}}\theta}}

(iii)

Out of syllabus

Full solutions

(i)

sin⁡(r+12)θ−sin⁡(r−12)θ≡sin⁡rθcos⁡12θ+cos⁡rθsin⁡12θ≡  −(sin⁡rθcos⁡12θ−cos⁡rθsin⁡12θ)≡2cos⁡rθsin⁡12θ  ■\begin{align*} &\sin \left({r + \textstyle \frac{1}{2}}\right) \theta - \sin \left({r - \textstyle \frac{1}{2}}\right) \theta \\ &\equiv \sin r \theta \cos {\textstyle \frac{1}{2}}\theta + \cos r \theta \sin {\textstyle \frac{1}{2}}\theta \\ &\phantom{\equiv} \; - \Big( \sin r \theta \cos {\textstyle \frac{1}{2}}\theta - \cos r \theta \sin {\textstyle \frac{1}{2}}\theta \Big) \\ &\equiv 2 \cos r \theta \sin {\textstyle \frac{1}{2}}\theta \; \blacksquare \end{align*}

(ii)

∑r=1ncos⁡rθ=∑r=1nsin⁡(r+12)θ−sin⁡(r−12)θ2sin⁡12θ=12sin⁡12θ(sin⁡32θ−sin⁡12θ+sin⁡52θ−sin⁡32θ+sin⁡72θ−sin⁡52θ⋯sin⁡(n−32)θ−sin⁡(n−52)θsin⁡(n−12)θ−sin⁡(n−32)θsin⁡(n+12)θ−sin⁡(n−12)θ)=sin⁡(n+12)θ−sin⁡12θ2sin⁡12θ  ■\begin{align*} &\sum_{r=1}^n \cos r \theta \\ &= \sum_{r=1}^n \frac{\sin \left({r + \textstyle \frac{1}{2}}\right) \theta - \sin \left({r - \textstyle \frac{1}{2}}\right) \theta }{2 \sin {\textstyle \frac{1}{2}}\theta} \\ & = \frac{1}{2 \sin {\textstyle \frac{1}{2}}\theta}\left( \def\arraystretch{1.5} \begin{array}{lclc} & \bcancel{\sin {\textstyle \frac{3}{2}\theta}} &-& \sin {\textstyle \frac{1}{2}\theta} \\ + & \bcancel{\sin {\textstyle \frac{5}{2}\theta}} &-& \bcancel{\sin {\textstyle \frac{3}{2}\theta}} \\ + & \bcancel{\sin {\textstyle \frac{7}{2}\theta}} &-& \bcancel{\sin {\textstyle \frac{5}{2}\theta}} \\ & & \cdots & \\ & \bcancel{\sin {\textstyle \left(n-\frac{3}{2}\right)\theta}} &-& \bcancel{\sin {\textstyle \left(n-\frac{5}{2}\right)\theta}} \\ & \bcancel{\sin {\textstyle \left(n-\frac{1}{2}\right)\theta}} &-& \bcancel{\sin {\textstyle \left(n-\frac{3}{2}\right)\theta}} \\ & \sin {\textstyle \left(n+\frac{1}{2}\right)\theta} &-& \bcancel{\sin {\textstyle \left(n-\frac{1}{2}\right)\theta}} \\ \end{array} \right) \\ &= \frac{\sin {\textstyle \left(n+\frac{1}{2}\right)\theta}-\sin {\textstyle \frac{1}{2}\theta}}{2 \sin {\textstyle \frac{1}{2}}\theta} \; \blacksquare \end{align*}

(iii)

Out of syllabus