Question

For triangle ABC prove that, cos (B+C/2)= sin A/2​

Answer 1

In Triangle ABC  

A°+B°+C°=180°  

A+B+C/2=180/2  

A+B/2=90-C/2  

MULTIPLY BY Sin Both Side  

Sin(A+B/2)=Sin(90-C/2)  

We Know That Sin(90-A)=CosA  

So,Sin(A+B/2)=CosC/2

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