यदि `A+B+C=pi` हो, तो सिद्ध कीजिए कि - `sin A + sin B + sin C = "4 cos"(A)/(2)"cos"(B)/(2)"cos"(C )/(2)`.

Explanation

`L.H.S. = sin A+sin B+sin C`
`="2 sin"(A+B)/(2)"cos"(A-B)/(2)+"2 sin"(C )/(2)"cos"(C )/(2)`
`= 2 sin(90^(@)-(C )/(2))"cos"(A-B)/(2)+"2 sin"(C )/(2)"cos"(C )/(2) " " [because (A+B+C)/(2)=90^(@)]`
`= "2 cos"(C )/(2)"cos"(A-B)/(2)+"2 sin"(C )/(2)"cos"(C )/(2)`
`="2 cos"(C )/(2)["cos"(A-B)/(2)+"sin"(C )/(2)]`
`= "2 cos"(C )/(2)["cos"(A-B)/(2)+sin (90^(@)-(A+B)/(2))]`
`= "2 cos"(C )/(2)["cos"(A-B)/(2)+"cos"(A+B)/(2)]`
`="2 cos"(C )/(2)"2 cos"(A)/(2)"cos"(B)/(2)`
`="4 cos"(A)/(2)"cos"(B)/(2)"cos"(C )/(2)=R.H.S. " "` इति सिद्धम्