Question

If A,B,C are interior angles of a triangle ABC,show that
Sin(B+C/2)=Sin A/2

Answer 1

Answer:

we know that , sum of interior angles of ∆ = 180°

A+B+C = 180°

B+C = 180-A

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

multiplying by sin both sides,

sin(B+C/2) = sin[90°-(A/2)]

sin(B+C/2) = CosA/2 (Ans)

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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....:)