Question

If $\cos A=\frac{3}{4}$, then $2\sin \frac{A}{2}\sin \frac{5A}{2}$

• A. $\frac{1}{16}$
• B. $-\frac{1}{16}$
• C. $-\frac{11}{16}$
• D. $\frac{11}{16}$

Answer 1

Answer: (D)

$\frac{11}{16}$

$\cos A=\frac{3}{4}$

$2\sin \frac{A}{2}\sin \frac{5A}{2}=\cos (\frac{A}{2}-\frac{5A}{2})$$-\cos (\frac{A}{2}+\frac{5A}{2})=\cos 2A-\cos 3A$

$\cos 2A=2{\cos }^{2}A-1=2{\left(\frac{3}{4}\right)}^2-1=\frac{1}{8}$

$\cos 3A=4{\cos }^{3}A-3\cos A=4{\left(\frac{3}{4}\right)}^3-3\left(\frac{3}{4}\right)=\frac{-9}{16}$

Hence $\cos 2A-\cos 3A=\frac{1}{8}-\frac{-9}{16}=\frac{11}{16}$.