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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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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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We 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 to

sin (B+C)/2 = cos A/2

Hence proved.

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