Question

If a,b and c are interior angles of a triangle abc. Then sin [(b+c)/2] = _____ *

Answer 1

Answer:

refer to the above attachment for solution

Answer 2

Answer:

If A, B, and C are interior angles of a triangle ABC. Then

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

Step-by-step explanation:

Let ABC be a triangle with interior angles A, B, C

we know that the sum of these angles is $A+B+C=180$

The value of $sin (\frac{B+C}{2} )$ can be calculated by rearranging the above equation

⇒

$A+B+C=180\\B+C=180-A\\\frac{B+C}{2}=\frac{180-A}{2}$

put the value of the angle

⇒$sin(\frac{B+C}{2})=sin(\frac{180-A}{2} )=sin(90-A/2)$

from trigonometry, we have the relation $sin(90-\alpha )=cos \alpha$

⇒$sin(\frac{B+C}{2} )=cos(A/2)$