If cot x= 5 12 for some x ∈ (π, 3 π 2 ), then sin 7 x (cos 13 x 2+sin 13 x 2 )+cos 7 x (cos 13 x 2-sin 13 x 2 ) is equal to

$$\cot x = \dfrac{5}{12} \Rightarrow \cos x = \dfrac{-5}{13} = 2\cos^2 \dfrac{x}{2} - 1$$cotx=125​⇒cosx=13−5​=2cos22x​−1

$\cos\left(\dfrac{x}{2}\right) = -\dfrac{2}{\sqrt{13}}$cos(2x​)=−13​2​ or $\dfrac{2}{\sqrt{13}}$13​2​ (rejected)

$$\left\{ \because \dfrac{x}{2} \in \left(\dfrac{\pi}{2}, \dfrac{3\pi}{4}\right) \right\}$${∵2x​∈(2π​,43π​)}

$$\left(\sin 7x \dfrac{\sin 13x}{2} + \cos 7x \dfrac{\cos 13x}{2}\right) + \left(\sin 7x \dfrac{\cos 13x}{2} - \cos 7x \dfrac{\sin 13x}{2}\right)$$(sin7x2sin13x​+cos7x2cos13x​)+(sin7x2cos13x​−cos7x2sin13x​)

$$\cos\left(7x - \dfrac{13x}{2}\right) + \sin\left(7x - \dfrac{13x}{2}\right)$$cos(7x−213x​)+sin(7x−213x​)

$$\cos \dfrac{x}{2} + \sin\left(\dfrac{x}{2}\right)$$cos2x​+sin(2x​)

$$\dfrac{3}{\sqrt{13}} - \dfrac{2}{\sqrt{13}} = \dfrac{1}{\sqrt{13}}$$13​3​−13​2​=13​1​