Question

If A, B and C are interior angles of a triangle ABC, then show that `sin((B+C)/2)=cosA/2`.

Answer 1

We know that the sum of the angles of a triangle is `180^@`.
`therefore A+B+C=180^@rArr B+C=180^@-A`
`rArr (B+C)/(2) =90^@-(A)/(2)`
`rArr sin ((B+C)/(2))=sin (90^@-(A)/(2))`
` rArr sin ((B+C)/(2))=cos.(A)/(2) [because sin (90^@-theta )=cos theta]`.
Hence,`sin ((B+C)/(2))=cos.(A)/(2)`.