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

Explanation

`L.H.S. ="cos"(A)/(2)+"cos"(B)/(2)+"cos"(C )/(2)`
`=2cos((A+B)/(4))cos((A-B)/(4))+sin((pi)/(2)-(C )/(2))`
`=2 cos((A+B)/(4))cos((A-B)/(4))+2sin((pi-C)/(4))cos((pi-C)/(4))`
`= 2 cos((pi-C)/(4))cos((A-B)/(4))+2sin((pi-C)/(4))cos((pi-C)/(4))`
`rArr L.H.S. = 2 cos((pi-C)/(4))["cos"(A-B)/(4)+sin((pi-C)/(4))]`
`=2cos((pi-C)/(4))["cos"(A-B)/(4)+sin((A+B)/(4))]`
`=2 cos((pi-C)/(4))[cos((A-B)/(4))+cos((pi)/(2)-(A+B)/(4))]`
`=cos((pi-C)/(4))["cos"(A-B)/(4)+cos((2pi-A-B)/(4))]`
`=2cos((pi-C)/(4))["cos"(2pi-A-B)/(4)+"cos"(A-B)/(4)]`
`=2cos((pi-C)/(4))[2cos((2pi-2B)/(2xx4))"cos"(2pi-2A)/(2xx4)]`
`=2 cos((pi-C)/(4))[2cos((pi-B)/(4))cos((pi-A)/(4))]`
`= 4cos((pi-A)/(4))cos((pi-B)/(4))cos((pi-C)/(4))`
`=R.H.S. " "` इति सिद्धम्