Question 0511

Sigma Notation
2011 Paper 1 Question 6 Variant

Question

(a)
Using the formulae for sin(A±B),{\sin(A \pm B),} prove that
sin(r+1)θsinr=2cos2r+12θsin12θ.\sin {\textstyle (r + 1)} \theta - \sin {\textstyle r} = 2 \cos { \textstyle \frac{2r+1}{2}} \theta \sin {\textstyle \frac{1}{2} \theta}.
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(b)
Hence find a formula for r=1ncos2r+12θ{\displaystyle \sum_{r=1}^n \cos { \textstyle \frac{2r+1}{2}} \theta} in terms of sin(n+1)θ,{\sin (n+1) \theta,} sinθ{\sin \theta} and sin12θ.{\sin \frac{1}{2}\theta.}
[3]

Answer