Let $\frac{\pi}{2}

Question

Let $\frac{\pi}{2} < x < \pi$ be such that $\cot x=\frac{-5}{\sqrt{11}}$. Then $\left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x)$ is equal to

Aicharya · Accepted Answer

Given, $\begin{aligned} & \cot x=-\frac{5}{\sqrt{11}} \ & \frac{1-	an ^2 \frac{x}{2}}{2 	an \frac{x}{2}}=-\frac{5}{\sqrt{11}} \ & 	an \frac{x}{2}=\sqrt{11},-\frac{1}{\sqrt{11}} \ & 	ext{So, } 	an \frac{x}{2}=\sqrt{11}, 	ext { As } \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}\end{aligned}$ $\begin{aligned} & 	ext{As, } x \in\left(\frac{\pi}{2}, \piight) \ & 	ext{Now, } \left(\sin \frac{11 x}{2}ight)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}ight)(\sin 6 x+\cos 6 x) \ & =\left\{\sin 6 x \sin \frac{11 x}{2}+\cos \frac{11 x}{2} \cos 6 x ight\} \ & =\cos \left(6 x-\frac{11 x}{2}ight)+\sin \left(6 x-\frac{11 x}{2}ight) \ & =\cos \frac{x}{2}+\sin \frac{x}{2} \ & =\frac{1}{2 \sqrt{3}}+\frac{\sqrt{11}}{2 \sqrt{3}} \ & =\frac{\sqrt{11}+1}{2 \sqrt{3}} \Rightarrow 	ext { Option (2) is correct. }\end{aligned}$

Let $\frac{\pi}{2} < x < \pi$ be such that $\cot x=\frac{-5}{\sqrt{11}}$. Then $\left(\sin \frac{11…

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