Let frac{pi}{2}

Question

(A) Given $\frac{\pi}{2} < 	heta < \pi$ and $\cot 	heta = -\frac{1}{2 \sqrt{2}}$.Expanding the expression: $\sin (\frac{15 	heta}{2}) \cos 8 	heta

Vedclass · Accepted Answer

$\frac{1 - \sqrt{2}}{\sqrt{3}}$

Vedclass · Answer

(A) Given $\frac{\pi}{2} Expanding the expression: $\sin (\frac{15 	heta}{2}) \cos 8 	heta + \sin (\frac{15 	heta}{2}) \sin 8 	heta + \cos (\frac{15 	heta}{2}) \cos 8 	heta - \cos (\frac{15 	heta}{2}) \sin 8 	heta$.Rearranging terms: $(\cos (\frac{15 	heta}{2}) \cos 8 	heta + \sin (\frac{15 	heta}{2}) \sin 8 	heta) + (\sin (\frac{15 	heta}{2}) \cos 8 	heta - \cos (\frac{15 	heta}{2}) \sin 8 	heta)$.Using identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$ and $\sin(A-B) = \sin A \cos B - \cos A \sin B$:$= \cos (8 	heta - \frac{15 	heta}{2}) + \sin (\frac{15 	heta}{2} - 8 	heta) = \cos (\frac{	heta}{2}) - \sin (\frac{	heta}{2})$.Since $\frac{\pi}{2}  \cos (\frac{	heta}{2})$, so $\cos (\frac{	heta}{2}) - \sin (\frac{	heta}{2}) Square the expression: $(\cos \frac{	heta}{2} - \sin \frac{	heta}{2})^2 = 1 - \sin 	heta$.Given $\cot 	heta = -\frac{1}{2 \sqrt{2}}$, $\sin 	heta = \frac{2 \sqrt{2}}{3}$.Thus, $(\cos \frac{	heta}{2} - \sin \frac{	heta}{2})^2 = 1 - \frac{2 \sqrt{2}}{3} = \frac{3 - 2 \sqrt{2}}{3} = \frac{(\sqrt{2} - 1)^2}{3}$.Taking the square root, since the expression is negative: $\cos \frac{	heta}{2} - \sin \frac{	heta}{2} = -\frac{\sqrt{2} - 1}{\sqrt{3}} = \frac{1 - \sqrt{2}}{\sqrt{3}}$.

Let $\frac{\pi}{2} < \theta < \pi$ and $\cot \theta = -\frac{1}{2 \sqrt{2}}$. Then the value of $\sin (\frac{15 \theta}{2}) (\cos 8 \theta + \sin 8 \theta) + \cos (\frac{15 \theta}{2}) (\cos 8 \theta - \sin 8 \theta)$ is equal to:

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