Question

if A,B,C are the interior angles of triangle ABC then show that Sin (B+ C)/2 =Cos A/2

Answer 1

since
A,B and C are the interior angles of the triangle,

then

A+B+C=π,

then

A/2 + B/2 + C/2 =π/2,

B/2 + C/2 = π/2 - A/2,

(B+C)/2 = (π/2 - A/2),

now taking sin in both sides, we have

sin(B+C)/2 = sin(π/2 - A/2),

sin(B+C)/2 =cosA/2,

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....:)