Trigonometry & Inverse Trigonometry
cot x Given — sin 7x·cos(13x/2) + cos 7x·sin(13x/2) Expression
nta_pyq_2026_jan
Grade None
Question:
If $\cot x=\dfrac{5}{12}$ for some $x\in\left(\pi,\dfrac{3\pi}{2}\right)$, then $\sin7x\left(\cos\dfrac{13x}{2}+\sin\dfrac{13x}{2}\right)+\cos7x\left(\cos\dfrac{13x}{2}-\sin\dfrac{13x}{2}\right)$ is equal to
\dfrac{4}{\sqrt{26}}
\dfrac{6}{\sqrt{26}}
\dfrac{5}{\sqrt{13}}
\dfrac{1}{\sqrt{13}}
Step-by-Step Solution
Key Concept: Regroup: $=\cos(7x-13x/2)+\sin(7x-13x/2)... = \cos(x/2)+\sin(x/2)$... wait from solution: expression $=\cos(x/2)+\sin(x/2)$ when correctly combined via sum formulas.
$\dfrac{1}{\sqrt{13}}$.
Correct Answer: 4