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!


closed as off-topic by Jonas Meyer, Mike Pierce, yoknapatawpha, Micah, user91500 Jul 18 '15 at 3:41

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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

  • $\begingroup$ How B+C becomes $\pi-A$? $\endgroup$ – Mathxx May 25 '15 at 15:24
  • 3
    $\begingroup$ note in a triangle ABC, $A+B+C=\pi=180^o$ $\endgroup$ – Harish Chandra Rajpoot May 25 '15 at 15:27

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