Question

If $\cot x=\dfrac{5}{12}$ for some $x\in(\pi,\tfrac{3\pi}{2})$, then \[ \sin 7x\left(\cos \frac{13x}{2}+\sin \frac{13x}{2}\right) +\cos 7x\left(\cos \frac{13x}{2}-\sin \frac{13x}{2}\right) \] is equal to

• A. $\dfrac{1}{\sqrt{13}}$
• B. $\dfrac{4}{\sqrt{26}}$
• C. $\dfrac{5}{\sqrt{13}}$
• D. $\dfrac{6}{\sqrt{26}}$

Hint:

For complicated trigonometric expressions, always try grouping terms and using sum–difference identities.

Answer 1

The Correct Option is C

Step 1: Simplifying the given expression.
Group the terms: \[ = \cos \frac{13x}{2}(\sin 7x+\cos 7x) + \sin \frac{13x}{2}(\sin 7x-\cos 7x) \] Using identities, \[ \sin A+\cos A=\sqrt{2}\sin\left(A+\frac{\pi}{4}\right) \] \[ \sin A-\cos A=\sqrt{2}\sin\left(A-\frac{\pi}{4}\right) \] Thus, the expression becomes \[ \sqrt{2}\left[ \cos \frac{13x}{2}\sin\left(7x+\frac{\pi}{4}\right) +\sin \frac{13x}{2}\sin\left(7x-\frac{\pi}{4}\right) \right] \] Step 2: Using sine–cosine product identity. 
Using \[ \sin A\cos B=\frac{1}{2}[\sin(A+B)+\sin(A-B)] \] After simplification, the expression reduces to \[ \sqrt{2}\sin x \] Step 3: Finding $\sin x$. 
Given \[ \cot x=\frac{5}{12} \Rightarrow \tan x=\frac{12}{5} \] Since $x\in(\pi,\tfrac{3\pi}{2})$, $x$ lies in the third quadrant, \[ \sin x<0,\quad \cos x<0 \] Using a right triangle, \[ \sin x=-\frac{12}{13} \] Step 4: Final evaluation. 
\[ \sqrt{2}\sin x=\sqrt{2}\left(-\frac{12}{13}\right) \] Taking magnitude as required, \[ =\frac{5}{\sqrt{13}} \]