If A is in the third quadrant and tan A = fra | vedclass

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(B) Given expression: $18 - 16 \sin^2 \frac{A}{2} - 32 \sin \frac{A}{2} \sin \frac{5A}{2}$ $= 18 - 8(2 \sin^2 \frac{A}{2}) - 16(2 \sin \frac{A}{2} \si

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(B) Given expression: $18 - 16 \sin^2 \frac{A}{2} - 32 \sin \frac{A}{2} \sin \frac{5A}{2}$ $= 18 - 8(2 \sin^2 \frac{A}{2}) - 16(2 \sin \frac{A}{2} \sin \frac{5A}{2})$ Using $2 \sin^2 	heta = 1 - \cos 2	heta$ and $2 \sin X \sin Y = \cos(X-Y) - \cos(X+Y)$: $= 18 - 8(1 - \cos A) - 16(\cos(2A) - \cos(3A))$ $= 18 - 8 + 8 \cos A - 16 \cos 2A + 16 \cos 3A$ $= 10 + 8 \cos A - 16(2 \cos^2 A - 1) + 16(4 \cos^3 A - 3 \cos A)$ Since $A$ is in the third quadrant and $	an A = \frac{\sqrt{7}}{3}$,we have $\cos A = -\frac{3}{4}$. Substituting $\cos A = -\frac{3}{4}$: $= 10 + 8(-\frac{3}{4}) - 16(2(\frac{9}{16}) - 1) + 16(4(-\frac{27}{64}) - 3(-\frac{3}{4}))$ $= 10 - 6 - 16(\frac{18}{16} - 1) + 16(-\frac{27}{16} + \frac{9}{4})$ $= 4 - 16(\frac{2}{16}) + 16(\frac{-27+36}{16})$ $= 4 - 2 + 9 = 11$.

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

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