Question

If `A+B+C=pi`, prove that : `cosA + cosB-cosC=4cos, A/2 cos, B/2 sin, C/2 -1

Answer 1

We have
`LHS=(cosA+cosB)-cos C`
`=2 cos ((A+B)/(2))cos((A-B)/(2))-(1-2 sin ^(2)""(C )/(2))`
`=2cos ((pi)/(2)=( C)/(2))cos ((A-B)/(2))-1+2sin^(2)""( C)/(2)`
`=2sin""( C)/(2)[cos((A-B)/(2))+sin ""(C )/(2)]-1`
` =2 sin ""( C)/(2)[ cos ((A-B)/(2))+sin {(pi)/(2)-((A+B))/(2)}]-1`
`= 2 sin ""(C )/(2)[ cos ((A-B)/(2))+ cos ((A+B)/(2))]-1`
`= 2 sin ""( C)/(2){2 cos ""(A)/(2)cos ""(B)/(2)}-1`
`={4 cos ""(B)/(2)sin ""( C)/(2)}-1=RHS.`
`therefore cos A+cos B-cosC=(4 cos ""(A)/(2)cos""(B)/(2)sin ""(C)/(2))-1.`