Question 0511

Sigma Notation
2011 Paper 1 Question 6 Variant

Question

(a)
Using the formulae for cos(A±B),{\cos(A \pm B),} prove that
cosrθcos(r+1)=2sin2r+12θsin12θ.\cos {\textstyle r} \theta - \cos {\textstyle (r + 1)} = 2 \sin { \textstyle \frac{2r+1}{2}} \theta \sin {\textstyle \frac{1}{2} \theta}.
[2]
(b)
Hence find a formula for r=1nsin2r+12θ{\displaystyle \sum_{r=1}^n \sin { \textstyle \frac{2r+1}{2}} \theta} in terms of cos(n+1)θ,{\cos (n+1) \theta,} cosθ{\cos \theta} and sin12θ.{\sin \frac{1}{2}\theta.}
[3]

Answer