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:

As we know, for any given triangle, the sum of all its interior angles is equals to 180°.

Thus,

A + B + C = 180° ….(1)

Now we can write the above equation as;

⇒ B + C = 180° – A

Dividing by 2 on both the sides;

⇒ (B + C)/2 = (180° – A)/2

⇒ (B + C)/2 = 90° – A/2

Now, put sin function on both sides.

⇒ sin (B + C)/2 = sin (90° – A/2)

Since,

sin (90° – A/2) = cos A/2

Therefore,

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

Answer 2

We will be using the trigonometric ratios of complementary angles to solve the given question.

sin (90° - θ) = cosθ

We know that for ΔABC,

∠A + ∠B + ∠C = 180° (Angle sum property of triangle)

∠B + ∠C = 180° - ∠A

On dividing both sides by 2, we get,

(∠B + ∠C)/2 = (180° - ∠A)/2

(∠B + ∠C)/2 = 90° - ∠A/2

Applying sine angles on both the sides:

sin {(∠B + ∠C)/2} = sin (90° - ∠A/2)

Since, sin (90° - θ) = cos θ, we get

sin (∠B + ∠C)/2 = cos A/2