Question

If $\cot x=\dfrac{5}{12}$ for some $x\in\left(\pi,\dfrac{3\pi}{2}\right)$, 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. $\dfrac{5}{\sqrt{13}}$
• C. $-\dfrac{1}{\sqrt{13}}$
• D. $\dfrac{8}{\sqrt{13}}$

Hint:

In trigonometric problems, always simplify long expressions first using sum-to-product or phase-shift identities.

Answer 1

The Correct Option is A

Step 1: Simplify the given expression.
\[ = \cos\frac{13x}{2}(\sin7x+\cos7x) + \sin\frac{13x}{2}(\sin7x-\cos7x) \] Step 2: Use standard identities.
\[ \sin7x+\cos7x=\sqrt{2}\cos\left(7x-\frac{\pi}{4}\right) \] \[ \sin7x-\cos7x=\sqrt{2}\sin\left(7x-\frac{\pi}{4}\right) \] Step 3: Combine the terms.
\[ = \sqrt{2}\cos\left(\frac{13x}{2}-\frac{\pi}{4}-7x\right) = \sqrt{2}\cos\left(\frac{-x}{2}-\frac{\pi}{4}\right) \] Step 4: Use $\cot x=\dfrac{5{12}$.}
\[ \sin x=-\frac{12}{13},\quad \cos x=-\frac{5}{13} \] (since $x$ lies in the third quadrant) Step 5: Evaluate the cosine term.
\[ \cos\left(\frac{x}{2}+\frac{\pi}{4}\right)=\frac{1}{\sqrt{26}} \] Step 6: Final value.
\[ \sqrt{2}\times\frac{1}{\sqrt{26}}=\frac{1}{\sqrt{13}} \]