Question

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

• A. $\frac{\sqrt{11}}{16}$
• B. $-\frac{\sqrt{11}}{16}$
• C. $\frac{11}{16}$
• D. $-\frac{5}{8}$

Answer 1

The correct option is D $-\frac{5}{8}$

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

${AC}^2={AB}^2+{BC}^2$

${BC}^2={AC}^2-{AB}^2$

$BC=\sqrt{16-9}=\sqrt{7}$

$2\sin \frac{A}{2}\sin \frac{5A}{2}$

$\cos C-\cos D=2\sin \left(\frac{D-C}{2}\right)\sin \left(\frac{D+C}{2}\right)$

$2\sin \left(\frac{D-C}{2}\right)\sin \left(\frac{D+C}{2}\right)=2\sin \frac{A}{2}\sin \frac{5A}{2}$

$\frac{D-C}{2}=\frac{A}{2};\frac{D+C}{2}=\frac{5A}{2}$

$D-C=A$

$D+C=5A$

$2D=6A$

$D=3A$

$D-C=A$

$C=D-A=2A$

$2\sin \frac{A}{2}\sin \frac{5A}{2}=\cos 2A-\cos A$

$=2{\cos }^{2}A-1-\cos A$

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

$=2\left(\frac{9}{16}\right)-1-\frac{3}{4}$

$=\frac{9}{8}-1-\frac{3}{4}=\frac{9-8-6}{8}$

$=\frac{-5}{8}$