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

Given △ABC

We know that sum of three angles of a triangle is 180

Hence ∠A+∠B+∠C=180°

or A+B+C=180°

B+C=180° −A

Multiply both sides by 1/2

1/2(B+C)= 1/2 (180° −A)

1/2(B+C)= 90°—A/2...(1)

Now

1/2(B+C)

Taking sine of this angle

sin( 2B+C)

[ 2B+C=90°−2A ]

sin(90°− 2A)

cos 2/A[sin(90°−θ)=cosθ]

Hence sin( 2B+C)=cos 2Aproved

Answer 2

Answer:

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

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Step-by-step explanation:

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