2011 H2 Mathematics Paper 1 Question 6

Sigma Notation

Answers

sin(n+12)θsin12θ2sin12θ{\frac{\sin {\left(n+\frac{1}{2}\right)\theta}-\sin { \frac{1}{2}\theta}}{2 \sin { \frac{1}{2}}\theta}}
Out of syllabus

Full solutions

(i)

sin(r+12)θsin(r12)θsinrθcos12θ+cosrθsin12θ  (sinrθcos12θcosrθsin12θ)2cosrθsin12θ  \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=1ncosrθ=r=1nsin(r+12)θsin(r12)θ2sin12θ=12sin12θ(sin32θsin12θ+sin52θsin32θ+sin72θsin52θsin(n32)θsin(n52)θsin(n12)θsin(n32)θsin(n+12)θsin(n12)θ)=sin(n+12)θsin12θ2sin12θ  \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