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+12)θ−sin⁡(r−12)=2cos⁡rθsin⁡12θ.\sin {\textstyle (r + \frac{1}{2})} \theta - \sin {\textstyle (r - \frac{1}{2})} = 2 \cos r \theta \sin {\textstyle \frac{1}{2} \theta}.
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(b)
Hence find a formula for ∑r=1ncos⁡rθ{\displaystyle \sum_{r=1}^n \cos r \theta} in terms of sin⁡(n+12)θ{\sin (n+\frac{1}{2}) \theta} and sin⁡12θ.{\sin \frac{1}{2}\theta.}
[3]

Answer