Question

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

• A. $-4$
• B. $-3$
• C. 3
• D. 4

Answer 1

Answer: (C)

$\cos A=\frac{3}{4}$
$16{\cos }^2\frac{A}{2}-32\sin \frac{A}{2}\sin \frac{5A}{2}$
$=\frac{16(1+\cos A)}{2}-16(\cos 2A-\cos 3A)$
$=\frac{16(1+\cos A)}{2}-16\left\{\left(2{\cos }^{2}A-1\right)-\left(4{\cos }^{3}A-3\cos A\right)\right\}$
$=8\left(1+\frac{3}{4}\right)-16\left\{2\times \frac{9}{16}-1-4\times \frac{27}{64}+3\times \frac{3}{4}\right\}=3$