Question

In a triangle ABC,if $A-B={120}^{\circ }$ and $\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=\frac{1}{32}$ then the value of $8\cos C$ is

Answer 1

$A-B=120andsin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=\frac{1}{32}....\left(1\right)$

$Weknowthat2sinAsinB=cos(A-B)-cos(A+B)$

$2sin\frac{A}{2}sin\frac{B}{2}=cos\left(\frac{A-B}{2}\right)-cos\left(\frac{A+B}{2}\right)....\left(2\right)$

$AndwealsoknowA+B+C=\pi SoA+B=\pi -C$

$So2sin\frac{A}{2}sin\frac{B}{2}=cos\left(\frac{120}{2}\right)-cos\left(\frac{\pi -C}{2}\right)=\frac{1}{2}-sin\frac{C}{2}....\left(3\right)$

$From\left(1\right)2sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=\frac{1}{16}$

$(\frac{1}{2}-sin\frac{C}{2})\left(sin\frac{C}{2}\right)=\frac{1}{16}$

$16si{n}^2\frac{C}{2}-8sin\frac{C}{2}+1=0$

$Sosin\frac{C}{2}=\frac{1}{4}$

$cos\frac{C}{2}=\frac{\sqrt{15}}{4}cosC=2co{s}^2\frac{C}{2}-1=\frac{2\times 15}{16}-1=\frac{7}{8}$

$So8cosC=7$