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If A+B+C=180^0, prove that : cos^2, A/2 ...

If `A+B+C=180^0`, prove that : `cos^2, A/2 + cos^2, B/2 - cos^2, C/2 = 2cos, A/2 cos, B/2 sin, C/2`

Text Solution

Verified by Experts

(a) `LHS=cos^(2)""(A)/(2)+cos^(2)""(B)/(2)-cos^(2)""(C)/(2)`
`=cos^(2)""(A)/(2)sin^(2)""(C)/(2)sin^(2)""(B)/(2)`
`=cos^(2)""(A)/(2)+sin((C-B)/(2))sin((C+B)/(2))`
`=cos^(2)""(A)/(2)+sin((C-B)/(2))cos((A)/(2))`
`=cos((A)/(2))[sin((B+C)/(2))+sin((C-B)/(2))]`
`=2cos((A)/(2))cos((B)/(2))sin((C)/(2))`
(b) `cos^(2)""(A)/(2)+cos^(2)""(B)/(2)+cos^(2)""(C)/(2)`
`=(1+cosA)/(2)+(1+cosB)/(2)+(1+cosC)/(2)`
`=(3+cosA+cosB+cosC)/(2)`
`=(3+1+4sin""(A)/(2)sin""(B)/(2)sin""(C)/(2))/(2)`
`=2+2sin""(A)/(2)sin""(B)/(2)sin""(C)/(2)`
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