Question

In a $△ABC$, If $A-B={120}^o$ and $\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)\sin \left(\frac{C}{2}\right)=\frac{1}{32}$, then the value of $8\cos C$ is

• A. $7$
• B. $8$
• C. $6$
• D. $9$

Answer 1

The correct option is A $7$

$\begin{array}{cc}\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=\frac{1}{32}& \\ \left(2\sin \frac{A}{2}\sin \frac{B}{2}\right)\sin \frac{C}{2}=\frac{2}{32}& |\begin{array}{c}A+B+C={180}^{\circ }\\ A+B={180}^{\circ }-C\\ \frac{A+B}{2}={90}^{\circ }-\frac{C}{2}\end{array}\\ \left(\cos \frac{A\_B}{2}-\cos \frac{A+B}{2}\right)\sin \frac{C}{2}=\frac{2}{32}& \\ \left[\cos {60}^{\circ }-\cos \left(\frac{A-B}{2}\right)\right]\sin \frac{C}{2}=\frac{1}{16}& \\ \left[\frac{1}{2}-\cos \left({90}^{\circ }-\frac{C}{2}\right)\right]\sin \frac{C}{2}=\frac{1}{16}& \\ \sin \frac{C}{2}\left(\frac{1}{2}-\sin \frac{C}{2}\right)=\frac{1}{16}& \\ \frac{1}{2}\sin \frac{C}{2}-{\sin }^2\frac{C}{2}=\frac{1}{16}& \\ {\sin }^2\frac{C}{2}-\frac{1}{2}-{\sin }^2\frac{C}{2}=\frac{1}{16}& \\ {\sin }^2\frac{C}{2}-\frac{1}{2}\sin \frac{C}{2}+{\left(\frac{1}{4}\right)}^2=0& \\ \left(\sin \frac{C}{2}-\frac{1}{4}\right)=0& \\ \Rightarrow \sin \frac{C}{2}=\frac{1}{4}& \\ \cos C=1-2{\sin }^2\frac{C}{2}=1-2\left(\frac{1}{16}\right)=1-\frac{1}{8}=\frac{7}{8}& \\ \therefore 8\cos C=7& \end{array}$