Trigonometry & Inverse Trigonometry
Simplification via Sum-to-Product
nta_pyq_2026_jan
Grade 11
Question:
Let $\dfrac{\pi}{2}<\theta<\pi$ and $\cot\theta=-\dfrac{1}{2\sqrt{2}}$. Then the value of $\sin\!\left(\dfrac{15\theta}{2}\right)(\cos8\theta+\sin8\theta)+\cos\!\left(\dfrac{15\theta}{2}\right)(\cos8\theta-\sin8\theta)$ is equal to
$-\dfrac{\sqrt{2}}{\sqrt{3}}$
$\dfrac{\sqrt{2}-1}{\sqrt{3}}$
$\dfrac{\sqrt{2}}{\sqrt{3}}$
$\dfrac{1-\sqrt{2}}{\sqrt{3}}$
Step-by-Step Solution
Key Concept: Regroup: $=\cos(8\theta-15\theta/2)+\sin(15\theta/2-8\theta)=\cos(\theta/2)+\sin(-\theta/2)=\cos(\theta/2)-\sin(\theta/2)$. Since $\pi/2<\theta<\pi$: $\pi/4<\theta/2<\pi/2$, so $\sin(\theta/2)>\cos(\theta/2)$.
$\dfrac{1-\sqrt{2}}{\sqrt{3}}$.
Correct Answer: 4