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If A, B and C are interior angles of a triangle ABC, then show that `sin((B+C)/2)=cos(A/2)`. |
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Answer» `R.H.S. = cos(A/2)` As we know, `A+B+C = 180^@` `A = 180-(B+C)` `:.cosA/2 = cos((180-(B+C))/2)=cos(90-(B+C)/2)`As we know, `cos(90-x) = sinx` `:.cosA/2 = sin((B+C)/2)=L.H.S.` |
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