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

In ΔABC

∠A + ∠B + ∠C = 180°  (angle sum property of triangles)

∴  ∠A = 180°- (∠B + ∠C  )         _________(i)

Now,

LHS = sin [(B + C)/2]

       = cos [ 90° - (B + C)/2]             [∵ sinФ = cos (90° - Ф)]

       = cos [ 180° - (B + C)]/2

       = cos A/2                                  [using (i)]

       = RHS

Hence Proved

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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....:)