Question

For ΔABC, prove that cos(B+C/2) = sin*A/2.

Answer 1

Hey there !

Solution:

In a Triangle ABC,

∠ A + ∠ B + ∠ C = 180° ( Angle Sum Property )

So using this equation, we get,

∠ B + ∠ C = 180 - ∠ A

Dividing by 2 on both sides we get,

=> ( ∠ B + ∠ C ) / 2 = 90 - ∠ A / 2

Applying Sin function on both sides, we get,

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

We know that,

Sin ( 90 - θ ) = Cos θ

Hence we can write,

Sin ( 90 - ∠ A / 2 ) = Cos ∠ A / 2

=> Sin ( ∠ B + ∠ C ) / 2 = Cos ∠ A / 2

Hence proved.

Hope my answer helped !

Answer 2

In ΔABC ,

∠A + ∠B + ∠C = 180° [ Angle sum property ]

⇒ B + C = 180° - A

⇒ ( B + C )/2 = 180°/2 - A/2

⇒ ( B + C )/2 = 90° - A/2

⇒ Cos ( B+C )/2 = Cos ( 90° - A/2 )

⇒ Cos ( B+C )/2 = Sin A/2

.....