Question

If A, B, C are angles of ∆ABC
Then show that,

sin{ B+C / 2 } = cos A/2​

Answer 1

Answer:

your answer is.

Step-by-step explanation:

A,B,C are the angles of triangle ABC

therefore, A+B+C=180

B+C=180-A

divide both sides with 2

B+C/2=180-A/2

multiply both sides with sin

sin{B+C/2}=sin{90-A/2}. (180÷2=90)

sin{B+C/2}=cos A/2. (sin(90-A)=cosA)

hence proved....

hope this helps you...