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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

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LHS=`cos^(2)A/2+cos^(2)B/2-cos^(2)C/2`
`=cos^(2)A/2+(1-sin^(2)B/2)-(1-sin^(2)C/2)`
`=cos^(2)A/2-sin^(2)B/2+sin^(2)C/2`
`=cosA/2-sin^(2)B/2+sin^(2)C/2`
`=cos(A+B)/(2).cos(A-B)/(2)+sin^(2)C2`
`=cos(180^(@)-C)/(2).cos(A-B)/(2) + sinC/2.sin(180^(@)-(A+2)/(2)`
`=sinC/2cos(A-B)/(2)+sinC/2cos(A+B)/(2)`
`=sinC/2[cos(A-B)/(2)+cos(A+B)/(2)]`
`=sinC/2.2cosA/2cosB/2`
`=2cosA/2cosB/2sinC/2`=RHS Hence Proved.
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