Question

please answer this question as soon as possible​

Answer 1

Given :-

If A,B and C are interior angles of triangle ABC

To Show :-

sin(B + C/2) = cos A/2

Solution :-

We know that sum of all angle in a triangle is 180

So,

∠A + ∠B + ∠C = 180

⇒ ∠B + ∠C = 180 - ∠A

⇒ B + C = 180 - A

On dividing both sides by 2

⇒ (B + C)/2 = (180 - A)/2

⇒ (B + C)/2 = 180/2 - A/2

⇒ (B + C)/2 = 90 - A/2

On multiplying both sides by sin

⇒ sin(B + C/2) = sin(90 - A/2)

we know that

• sin(90 - A/2) = cos A/2

⇒ sin(B + C/2) = cos A/2

Hence, proved

Answer 2

Hope this helps u sorry for putting this in ur question