Trigonometry & Inverse Trigonometry
Trigonometric Ratios and Identities
nta_pyq_2025_apr
Grade 11
Question:
If $\displaystyle\sum_{r=1}^{13} \left\{\frac{1}{\sin\!\left(\tfrac{\pi}{4}+(r-1)\tfrac{\pi}{6}\right)\sin\!\left(\tfrac{\pi}{4}+r\tfrac{\pi}{6}\right)}\right\} = a\sqrt{3} + b,\ a, b \in \mathbf{Z}$, then $a^2 + b^2$ is equal to
Step-by-Step Solution
Key Concept: Multiply and divide by $\sin(\pi/6)=1/2$ and use $\sin(B-A)=\sin B\cos A-\cos B\sin A$ to telescope each term as $\cot\!\left(\tfrac{\pi}{4}+(r-1)\tfrac{\pi}{6}\right)-\cot\!\left(\tfrac{\pi}{4}+r\tfrac{\pi}{6}\right)$.
Each term $=\dfrac{1}{\sin(\pi/6)}\left[\cot\!\left(\tfrac{\pi}{4}+(r-1)\tfrac{\pi}{6}\right)-\cot\!\left(\tfrac{\pi}{4}+r\tfrac{\pi}{6}\right)\right]$ (standard telescoping). Summing from $r=1$ to $13$: $=2\left[\cot\tfrac{\pi}{4}-\cot\left(\tfrac{\pi}{4}+\tfrac{13\pi}{6}\right)\right]=2\left[1-\cot\left(\tfrac{29\pi}{12}\right)\right]$. Now $\tfrac{29\pi}{12}=2\pi+\tfrac{5\pi}{12}$, so $\cot\tfrac{29\pi}{12}=\cot\tfrac{5\pi}{12}=\cot75°=2-\sqrt{3}$. Sum $=2[1-(2-\sqrt{3})]=2\sqrt{3}-2=a\sqrt{3}+b$. Hence $a=2$, $b=-2$, $a^2+b^2=4+4=8$.
Correct Answer: 4