Question:
In $\Delta A B C$ if $\cos C=\frac{\sin A}{2 \sin B}$, prove that the triangle is isosceles.
Solution:
Let $\Delta A B C$ be any triangle.
Suppose $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k$
If $\cos C=\frac{\sin A}{2 \sin B}$, then
$\frac{b^{2}+a^{2}-c^{2}}{2 a b}=\frac{k a}{2 k b} \quad\left(\because \cos C=\frac{b^{2}+a^{2}-c^{2}}{2 a b}\right)$
$\Rightarrow b^{2}+a^{2}-c^{2}=a^{2}$
$\Rightarrow b^{2}-c^{2}=0$
$\Rightarrow(b-c)(b+c)=0$
$\Rightarrow b-c=0$
$\Rightarrow b=c \quad(\because b, c>0)$
Thus, the lengths of two sides of the $\Delta A B C$ are equal.
Hence, $\Delta \mathrm{ABC}$ is an isosceles triangle.