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In triangle ABC, prove that cos(A/2)+cos...

In triangle ABC, prove that `cos(A/2)+cos(B/2)+cos(C/2)=4cos(pi-A)/4cos(pi-B)/4cos(pi-C)/4`

Text Solution

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`R.H.S. = 4cos((pi-A)/4)cos((pi-B)/4)cos((pi-C)/4)`
`=2*(2cos((pi-A)/4)cos((pi-B)/4))cos((pi-C)/4)`
`=2*(cos((pi-A+pi-B)/4)+cos((pi-A-pi+B)/4))cos((pi-C)/4)`
`=2*(cos(pi/2-(A+B)/4)+cos((B-A)/4))cos((pi-C)/4)`
`=2*(sin((A+B)/4)+cos((B-A)/4))cos((A+B)/4)`....(As `a+b+c = pi`)
`= (2sin((A+B)/4)cos((A+B)/4))+(2cos((B-A)/4)cos((A+B)/4))`
`=sin((A+B)/2)+cos((B-A+A+B)/4)+cos((B-A-A-B)/4)`
`=sin((pi-C)/2)+cos(B/2)+cos(-A/2)`
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