Sin(B+C/2)=cos(A/2). Prove

1.	Sin(B+C/2)=cos(A/2). Prove
Answer» A+B+C/2=180/2B+C/2=90=-A/2B+C/2=90-A/2multiplying both sides by sinA we getsin(B+C/2)=sin(90-A/2)sin(B+C/2)=cos(A/2)hence it is proved A+B+C/2=180/2 In triangle abc A+B+C=180dividing both side by 2A/2 +B/2+c/2=90b/2+c/2=90-a/2sin(b+c/2)=sin(90-a/2)sin(b+c/2)=cot(a/2)

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