Question

Hello brainlies!!!

Question :
If A, B, C are the angles of a triangle ABC show that
$\sin( \frac{b + c}{2} ) = \cos( \frac{a}{2} )$
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Answer 1

Hello dear❤️

Solution:

We know that the sum of all angles of a

triangle is 180°

A+B+C = 180° = B+C = 180° - A

$\frac{b + c}{2} = {90} \: degree - \frac{a}{2}$

$\sin( \frac{b + c}{2} ) = \sin(90 \: degree - \frac{a}{2} )$

$\sin( \frac{b + c}{2} ) = \cos( \frac{a}{2} )$

Hence,

$\sin( \frac{b + c}{2} ) = \cos( \frac{a}{2} )$

Hope this helps dear ❤️

Answer 2

Step-by-step explanation:

angle A + Angle B + Angle C = 180°

Angle B + Angle C = 180°-A

Then,

Angle B + Angle C/2 = 90°-A/2

Therefore,

Sin(B+C/2) = Sin(90°-A/2)

= Sin(B+C/2) = Cos A/2 [ Sin(90°-theta) = cos theta]

Hence,

LHS = RHS

Sin (B+C/2) = CosA/2