Question

If A, B and C are interior angles of a triangle ABC, then \[\sin \left( \frac{B + C}{2} \right) =\]

Options

• \[\sin \frac{A}{2}\]

• \[\cos \frac{A}{2}\]

• \[- \sin \frac{A}{2}\]

• \[- \cos \frac{A}{2}\]

Answer 1

We know that in triangle `ABC`

`A+B+C=180°`

⇒ `B+C=180°-A` 

⇒` (B+C)/2=(90°)/2-A/2` 

⇒ `sin ((B+C)/2)=sin (90°-A/2)`

`"since" sin (90°-A)=cos A` 

So 

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