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:

Step-by-step explanation:

A + B + C = 180

=> B + C = 180 - A

Divide by 2 on both sides

=> (B + C)/2 = (180 - A)/ 2

=> (B + C) /2 = 90 - (A/2)

Take Sine on both sides

=> Sin((B+C)/2) = Sin((90 - A/2))

=> Sin((B+C)/2)  = Cos A/2 (∵ Sin (90 - θ) = Cosθ)

Hence proved.

Answer 2

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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

Hope it's Helpful....:)