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

Dear Student

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.

Thanks