Question

If A, B and C are interior angles of a triangle ABC, then $sin(\frac{B+C}{2} )$=
(a$sin(\frac{A}{2} )$
(b)$cos(\frac{A}{2} )$
(c)$-sin(\frac{A}{2} )$
(d)$-cos(\frac{A}{2} )$

Answer 1

SOLUTION :  

The correct option is (b) : cosA/2

Given : If A, B, C are the interior angles of a ∆ ABC.

In ∆ ABC,

A + B + C=180°

[Sum of the angles of a ∆ is 180°]

B + C = 180° - A

(B+ C)/2 = 180° /2 − A/2

[On Dividing both Sides by 2]

(B+ C)/2 = 90° /2 −A/2

On taking sin both sides,

sin (B+C)/2 = sin (90° - A)/2

sin(B+C)/2 = cosA/2         [sin (90 - θ) = cos θ]

Hence , sin(B+C)/2 = cosA/2 .  

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

the correct option is b.