Question

If A, B and Care interior angles of a triangle ABC, then show that sin(b+c/2)= cos a/2
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Answer 1

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If a, b and c are the interior angles of a triangle, then

a+b+c=180°(Angle sum property of a triangle)

b+c=180°-a

Dividing by 2 on both sides,

b+c/2=180°-a/2

Taking sin on both the sides,

sin(b+c/2)=sin(180°-a/2)

sin(b+c/2)=sin(90°-a/2)

sin(b+c/2)=cos(a/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....:)