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If A, B and C are interior angles of a triangle ABC, then show that sin((B+C)/2) = cosA/2

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In a triangle, sum of all the interior angles

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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