Trigonometric Ratios and Identities

Concept:Simplify the given trigonometric expression using sum-to-product identities and then evaluate using the given $\cot x$ value and quadrant information.Explanation:Given $\cot x = \frac{5}{12}$ and $x \in (\pi, \frac{3\pi}{2})$ (third quadrant), so $\sin x$ is negative.The expression is: $\sin 7x\left(\cos\frac{13x}{2}+\sin\frac{13x}{2}\right) + \cos 7x\left(\cos\frac{13x}{2}-\sin\frac{13x}{2}\right)$.Expand the brackets: $\sin7x\cos\frac{13x}{2} + \sin7x\sin\frac{13x}{2} + \cos7x\cos\frac{13x}{2} - \sin\frac{13x}{2}\cos7x$.Rearrange: $(\sin7x\cos\frac{13x}{2} - \sin\frac{13x}{2}\cos7x) + (\cos7x\cos\frac{13x}{2} + \sin7x\sin\frac{13x}{2})$.Use identities: $\sin A\cos B - \cos A\sin B = \sin(A-B)$ and $\cos A\cos B + \sin A\sin B = \cos(A-B)$.So the expression becomes: $\sin\left(7x - \frac{13x}{2}\right) + \cos\left(7x - \frac{13x}{2}\right) = \sin\frac{x}{2} + \cos\frac{x}{2}$.Since $x \in (\pi, \frac{3\pi}{2})$, then $\frac{x}{2} \in (\frac{\pi}{2}, \frac{3\pi}{4})$ (second quadrant). There $\sin\frac{x}{2}>0$, $\cos\frac{x}{2}<0$, but $\sin\frac{x}{2} > |\cos\frac{x}{2}|$, so $\sin\frac{x}{2}+\cos\frac{x}{2}>0$.Let $L = \sin\frac{x}{2}+\cos\frac{x}{2}$. Square: $L^2 = \sin^2\frac{x}{2}+\cos^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2} = 1+\sin x$.We have $\cot x = \frac{5}{12}$, so $\csc x = -\sqrt{1+\cot^2 x} = -\sqrt{1+\frac{25}{144}} = -\sqrt{\frac{169}{144}} = -\frac{13}{12}$ (negative in third quadrant).Thus $\sin x = \frac{1}{\csc x} = -\frac{12}{13}$.Hence $L^2 = 1 + \left(-\frac{12}{13}\right) = \frac{1}{13}$. Taking positive root: $L = \frac{1}{\sqrt{13}}$.Answer:$\frac{1}{\sqrt{13}}$ which corresponds to option D.