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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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Answer» Answer 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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