Question

If A, B and C are interior angles of a ΔABC then cos $\sf{(\frac{B+C}{2})}$is equal to ?​

Answer 1

Solution :-

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 this might help you mate..

Answer 2

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Answer

➜ ∠A+∠B+∠C=180°[Angle sum

property of a triangle]

➜ ∠B+∠C=180−∠A

➜ ∠B+∠C / 2 = 90° − ∠A / 2

➜ sin ( B+C / 2 ) = sin( 90 − A / 2)

➜ cos A / 2

Hence Proved.