Question

prove that sin (A+C)/2 =cos B/2 ​

Answer 1

EXPLANATION.

Prove that : sin(A + C)/2 = cos(B/2).

As we know that,

Sum of the angles of a triangle = 180°.

⇒ A + B + C = 180°.

⇒ A + C = 180° - B.

Divide both sides by 2, we get.

⇒ (A + C)/2 = (180° - B)/2.

⇒ (A + C)/2 = [90° - (B/2)].

Now we can write equation as,

⇒ sin(A + C)/2.

Put the values in the equation, we get.

⇒ sin[90° - (B/2)].

As we know that,

⇒ sin(90° - θ) = cosθ.

Using this identities in the equation, we get.

⇒ cos(B/2).

Hence Proved.