Question

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

Sin[B+C/2]=Sin[A/2]
Answer ASAP

Answer 1

Answer:

sin[(B+C)/2] = cos(A/2)  

Step-by-step explanation:

Given, A, B and C are interior angles of a triangle ABC

=> A+B+C = 180°

=> B+C = 180-A

=> (B+C)/2 = (180-A)/2

=> (B+C)/2 = 90 - (A/2)

=> sin[(B+C)/2] = sin (90 - (A/2) )

=> sin[(B+C)/2] = cos(A/2)  

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