Question

If A,B and C are interior angles of triangle ABC, then prove that sin(B+C/2) = cos A/2​

Answer 1

Answer:

in triangle

A + B + C = 180

B + C = 180 - A

sin((B+C)/2) = cos A/2

LHS = sin((180-A)/2)

= sin(90-A/2). we know sin(90-∅) = cos∅

= cosA/2

LHS = RHS

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....:)