The question with the options
Q: $If\ \cos \ A\ =\ \frac{3}{4},\ then\ the\ value\ of\ 16\cos ^2\left(\frac{A}{2}\right)-32\sin \left(\frac{A}{2}\right)\ \sin \left(\frac{5A}{2}\right)\ is$
$\left(a\right)\ -4\ \ \left(b\right)\ -3\ \ \left(c\right)\ 3\ \ \left(d\right)\ \ $
Answer: 3, Option C
Solution:
$Given,\ \cos \ A\ =\ \frac{3}{4}$
$and\ 16\cos ^2\left(\frac{A}{2}\right)-32\sin \left(\frac{A}{2}\right)\ \sin \left(\frac{5A}{2}\right)$
$Formula\ used:\ \cos 2A=2\cos ^2A−1$
$Formula\ used:\ \cos 3A=4\cos ^3A−3\cos A$
$\cos A=\ 2\cos ^2\left(\frac{A}{2}\right)-1$
$2\ \cos ^2\left(\frac{A}{2}\right)\ =\ \cos A+1$
$=\ \frac{3}{4}+1$
$=\frac{7}{4}$
$\cos ^2\left(\frac{A}{2}\right)=\frac{7}{8}$
$and\ 16\cos ^2\left(\frac{A}{2}\right)-32\sin \left(\frac{A}{2}\right)\ \sin \left(\frac{5A}{2}\right)$
$16\cdot \frac{7}{8}-16\left(2\ \sin \frac{A}{2}.\sin \frac{5A}{2}\right)$
$14-16\left(2\cos 2A-\cos 3A\right)$
$14-16\left[\left(2\cos ^2A-1\right)-\left(4\cos ^3A-3\cos A\right)\right]$
$14-16\left[\left(2\cdot \frac{9}{16}-1\right)-4\left(-\frac{9}{16}\right)\right]$
$14-2-9$
$=3$