Let $\frac{\pi}{2}

Question

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

Professor PC Thomas Classes · Answer

(b) : $\cot x=-\frac{5}{\sqrt{11}}, \frac{\pi}{2}<x<\pi$ Let $E=\sin \frac{11 x}{2}(\sin 6 x-\cos 6 x)+\cos \frac{11 x}{2}$ $$ (\sin 6 x+\cos 6 x) $$ $=\sin \frac{11 x}{2} \cdot \sin 6 x+\cos \frac{11 x}{2} \cdot \cos 6 x$ $$ +\sin 6 x \cdot \cos \frac{11 x}{2}-\cos 6 x \cdot \sin \frac{11 x}{2} $$ $=\cos \left(6 x-\frac{11 x}{2}ight)+\sin \left(6 x-\frac{11 x}{2}ight)=\cos \frac{x}{2}+\sin \frac{x}{2}$\ Now, $\cot x=\frac{-5}{\sqrt{11}} \Rightarrow \frac{1-	an ^{2} \frac{x}{2}}{2 	an \frac{x}{2}}=\frac{-5}{\sqrt{11}}$\ $\Rightarrow \quad 	an ^{2} \frac{x}{2}-\frac{10}{\sqrt{11}} 	an \frac{x}{2}-1=0$\ giving $	an \frac{x}{2}=\sqrt{11},-\frac{1}{\sqrt{11}}$\ $	herefore \quad 	an \frac{x}{2}=\sqrt{11} ;$ as $\frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}$\