If $A$ is in the third quadrant and $ an A=\frac{\sqrt{7}}{3}$, then $18-16 \sin ^2 \frac{A}{2}-32 \sin \frac{A}{2} \sin \frac{5 A}{2}=$

Question

If $A$ is in the third quadrant and $	an A=\frac{\sqrt{7}}{3}$, then $18-16 \sin ^2 \frac{A}{2}-32 \sin \frac{A}{2} \sin \frac{5 A}{2}=$

Aicharya · Accepted Answer

$ \begin{aligned} & 18-16 \sin ^2 \frac{A}{2}-32 \sin A / 2 \sin 5 A / 2 \\ & =18-8\left(2 \sin ^2 \frac{A}{2}\right)-16\left(2 \sin \frac{A}{2} \sin \frac{5 A}{2}\right) \\ & =18-8(1-\cos A)-16\left(\cos \left(\frac{5 A}{2}-\frac{A}{2}\right)-\cos \left(\frac{5 A}{2}+\frac{A}{2}\right)\right) \\ & \quad \quad[\operatorname{ascos}(A-B)-\cos (A+B)=2 \sin A \sin B \\ & =18-8+8 \cos A-16 \cos 2 A+16 \cos 3 A \\ & =10+8 \cos A-16\left[2 \cos ^2 A-1\right] \\ & \quad+16\left[4 \cos ^3 A-3 \cos 2 A\right] \end{aligned} $ [as $\cos 3 A=4 \cos ^3 A-3 \cos A, \cos 2 A=2 \cos ^2 A-1$ ] Now, $A$ lies in third quadrant and $\tan A=\frac{\sqrt{7}}{3}$

$\begin{aligned} & \cos A=\frac{-3}{4} \\ & =10+8\left(\frac{-3}{4}\right)-16\left[2 \cdot\left(\frac{9}{16}\right)-1\right] \\ & +16\left[4 \cdot\left(\frac{-27}{64}\right)-3 \cdot\left(\frac{-3}{4}\right)\right]\end{aligned}$ $\begin{aligned} & =10-6-16 \cdot \frac{1}{8}+16\left(\frac{-27}{16}+\frac{9}{4}\right) \\ & =10-6-2+16 \cdot \frac{9}{16}=2+9=11 .\end{aligned}$

If $A$ is in the third quadrant and $\tan A=\frac{\sqrt{7}}{3}$, then $18-16 \sin ^2 \frac{A}{2}-32 \sin…

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