6.    If A,B and C are interior angles of a triangle ABC, then show that

          \sin (\frac{B+C}{2})= \cos \frac{A}{2}

        

Answers (1)
M manish

Given that,
A, B and C are interior angles of \Delta ABC 
To prove - \sin (\frac{B+C}{2})= \cos \frac{A}{2}

Now,
In triangle \Delta ABC
       A + B + C  = 180^0
\Rightarrow B + C = 180 - A
\Rightarrow B + C/2 = 90^0 - A/2
\sin \frac{B+C}{2}=\sin (90^0-A/2)
\sin \frac{B+C}{2}=\cos A/2
Hence proved 
 

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