<s>[INST] Prove this is an isosceles triangle In a $\triangle ABC$, $$\qquad \sin B\cdot\sin C=\cos^2\left(\frac{A}{2}\right)$$
Prove that this is an isosceles triangle. 
Can anyone guide me to prove this? Thanks!
 [/INST] Note: in a triangle ABC, we have $A+B+C=\pi=180^o$
Given $$\sin B\sin C=\cos^2\frac{A}{2} \implies 2\sin B\sin C=2\cos^2\frac{A}{2} $$$$\implies \cos(B-C)-\cos(B+C)=2\cos^2\frac{A}{2} $$$$\implies \cos(B-C)-\cos(180^o-A)=2\cos^2\frac{A}{2} $$ $$\implies \cos(B-C)+\cos A=2\cos^2\frac{A}{2} $$$$\implies \cos(B-C)+2\cos^2\frac{A}{2}-1=2\cos^2\frac{A}{2}$$ $$\cos(B-C)=1 \implies B-C=0 \quad \text{or} \quad  B=C$$ Hence, the triangle ABC is an isosceles triangle
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