A trigonometric block sum · source reference

Printed question

Given $\sum_{r=1}^n\cos2r\theta=\frac{\sin(2n+1)\theta-\sin\theta}{2\sin\theta}$, where $0<\theta<\pi$, show that

$$\sin^2 10\theta+\sin^2 11\theta+\cdots+\sin^2 20\theta=k-\frac{\sin41\theta-\sin19\theta}{4\sin\theta},$$

where $k$ is a constant to be determined. [4]

Printed working

$$\begin{aligned}\sum_{r=10}^{20}\sin^2r\theta&=\sum_{r=10}^{20}\frac{1-\cos2r\theta}2\\&=\frac12(20-10+1)-\frac12\left[\sum_{r=1}^{20}\cos2r\theta-\sum_{r=1}^9\cos2r\theta\right]\\&=\frac{11}2-\frac12\left[\frac{\sin41\theta-\sin\theta}{2\sin\theta}-\frac{\sin19\theta-\sin\theta}{2\sin\theta}\right]\\&=\frac{11}2-\frac{\sin41\theta-\sin19\theta}{4\sin\theta}.\end{aligned}$$

Thus $k=\frac{11}2$.

Teacher notes

Formula found in list MF27: $\sin^2x=\frac{1-\cos2x}2$.

Printed answer key

$k=11/2$

Mathematical review and qualifications

$\sum_{r=10}^{20}\sin^2(r\theta)=\frac{11}2-\frac12\sum_{r=10}^{20}\cos(2r\theta)$. The cosine block equals $[\sin(41\theta)-\sin(19\theta)]/(2\sin\theta)$. Therefore the requested expression holds with $k=11/2$. Since $0<\theta<\pi$, its denominator is nonzero.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: SIGMA NOTATION. [modified 2024/RI/Prelim/I/5] Question PDF pages 1; solution PDF pages 4. The original files remain private QA records outside this native content.