Question

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

Answer 1

Answer: we know the formula of Cos theta (sin90-sin theta) so,

We have sin ( B+C/2)

then sin (90-B+C/2)= cos A/2

90-B-C/2=A/2

= 90=A/2+B+C/2

=90= A+B+C/2

=A+B+C=180(we know the sum of interior angles of a triangle )

B+C=180-A

B+C/2=90-A/2(divided by 2)

Sin(B+C/2)=Sin(90-90+A/2)

Sin(B+C)=CosA/2 prooved

Step-by-step explanation: