Question

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

Answer 1

To prove:

$\dfrac{sin(B+C)}{2} = cos\dfrac{A}{2}$

Proof:

L.H.S = $\dfrac{sin(B+C)}{2}$

R.H.S = $cos\dfrac{A}{2}$

We know,

$\angle A + \angle B + \angle C = 180^{\circ}$ (Angle Sum Property)

$\angle B + \angle C = 180^{\circ} - A$

R.H.S. = $cos\dfrac{A}{2}$ $=sin(90^{\circ} - \dfrac{A}{2})$

L.H.S. = $sin\dfrac{(B+C)}{2}$

$= sin \dfrac{(180^{\circ}- A)}{2}$

$= sin(90^{\circ}-\dfrac{A}{2})$

LHS = RHS

Hence, proved.

Identity used:

$cos\theta = sin (90^{\circ}- \theta)$

Answer 2

Answer:

Hello mate

here's your answer

please mark me as the brainlist