Let frac{pi}{2}

Question

(B) Given expression: $E = \sin \frac{11x}{2} \sin 6x - \sin \frac{11x}{2} \cos 6x + \cos \frac{11x}{2} \sin 6x + \cos \frac{11x}{2} \cos 6x$$= (\cos

Vedclass · Accepted Answer

$\frac{\sqrt{11}+1}{2\sqrt{3}}$

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(B) Given expression: $E = \sin \frac{11x}{2} \sin 6x - \sin \frac{11x}{2} \cos 6x + \cos \frac{11x}{2} \sin 6x + \cos \frac{11x}{2} \cos 6x$$= (\cos \frac{11x}{2} \cos 6x + \sin \frac{11x}{2} \sin 6x) + (\cos \frac{11x}{2} \sin 6x - \sin \frac{11x}{2} \cos 6x)$$= \cos(6x - \frac{11x}{2}) + \sin(6x - \frac{11x}{2})$$= \cos \frac{x}{2} + \sin \frac{x}{2}$Since $\cot x = \frac{-5}{\sqrt{11}}$,we have $\frac{1 - 	an^2(x/2)}{2 	an(x/2)} = \frac{-5}{\sqrt{11}}$.Let $t = 	an(x/2)$. Then $\sqrt{11}(1 - t^2) = -10t \implies \sqrt{11}t^2 - 10t - \sqrt{11} = 0$.Solving for $t$: $t = \frac{10 \pm \sqrt{100 - 4(\sqrt{11})(-\sqrt{11})}}{2\sqrt{11}} = \frac{10 \pm \sqrt{144}}{2\sqrt{11}} = \frac{10 \pm 12}{2\sqrt{11}}$.Since $\frac{\pi}{2}  0$. Thus $t = \frac{22}{2\sqrt{11}} = \sqrt{11}$.Using $	an(x/2) = \sqrt{11}$,we find $\sin(x/2) = \frac{\sqrt{11}}{\sqrt{12}} = \frac{\sqrt{11}}{2\sqrt{3}}$ and $\cos(x/2) = \frac{1}{\sqrt{12}} = \frac{1}{2\sqrt{3}}$.Therefore,$E = \frac{1}{2\sqrt{3}} + \frac{\sqrt{11}}{2\sqrt{3}} = \frac{\sqrt{11}+1}{2\sqrt{3}}$.

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

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