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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
Dear StudentWe know that, for a given triangle,sum of all the interior angles of a triangle is equal to 180°A + B + C = 180°....(1)To find the value of (B+ C)/2, simplify the equation (1)⇒B + C = 180° - A⇒(B+C)/2 = (180°-A)/2⇒(B+C)/2 = (90°-A/2)Now, multiply both sides by sin functions,we get⇒sin (B+C)/2 = sin (90°-A/2)Since sin (90°-A/2) = cos A/2,the above equation is equal tosin (B+C)/2 = cos A/2Hence proved.Thanks
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