Question

For any triangle ABC, if $B=3C$, show that $\cos C=\sqrt{\frac{b+c}{4c}}$ and $\sin \frac{A}{2}=\frac{b-c}{2c}$.

Answer 1

Given,

$B=3C$

Taking $\sin$ on both sides, we have

$\sin B=\sin 3C$

$\sin B=3\sin C-4{\sin }^{3}C$

$=\sin C(3-4{\sin }^{2}C)$

$=\sin C(4{\cos }^{2}C-1)$

$\sin B=\sin C(4{\cos }^{2}C-1)$

$\frac{\sin B}{\sin C}={\cos }^{2}C-1$ ... (1)

Using $\sin$ formula, we have

$\frac{\sin B}{b}=\frac{\sin C}{c}$

$\frac{\sin B}{\sin C}=\frac{b}{c}$ ...... (2)

From equation (1) and (2),

$4{\cos }^{2}C-1=\frac{b}{c}$

$4{\cos }^{2}C=\frac{b}{c}+1$

$\cos C=\sqrt{\frac{b+c}{4c}}$ ...(3)

Now, in a tranagle taking $A+B+C={180}^{\circ }$

$90-\frac{A}{2}=\frac{B+C}{2}$

$\cos (90-\frac{A}{2})=\cos \left(\frac{B+C}{2}\right)$

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

Since, $B=3C$, we have

$\sin \frac{A}{2}=cos\frac{3C+C}{2}$

$\sin \frac{A}{2}=\cos 2C$

$\sin \frac{A}{2}=2{\cos }^{2}C-1$ ...(4)

From equation (3) and (4)

$\sin \frac{A}{2}=\frac{b+c}{2c}-1$

$\sin \frac{A}{2}=\frac{b-c}{2c}$

Hence proved.