If A,B,C are the angles of a triangle then cosA+cosB+cosC=?

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Answer 1 · 2017-11-02T16:51:11

$cos A + cos B + cos C = 1 + 4 sin (\frac{A}{2}) sin (\frac{B}{2}) sin (\frac{C}{2})$ $cosA+cosB+cosC=1+4sin(A/2)sin(B/2)sin(C/2)$

As $A, B$ $A,B$ and $C$ $C$ are angles of a triangle, we have

$A + B + C = 180$ $A+B+C=180$ and $A + B = 180 - C$ $A+B=180-C$ or $\frac{A + B}{2} = 90 - \frac{C}{2}$ $(A+B)/2=90-C/2$

Hence $cos A + cos B + cos C$ $cosA+cosB+cosC$

= $2 cos (\frac{A + B}{2}) cos (\frac{A - B}{2}) + 1 - 2 {sin}^{2} (\frac{C}{2}) - 1$ $2cos((A+B)/2)cos((A-B)/2)+1-2sin^2(C/2)-1$

= $2 cos (90 - \frac{C}{2}) cos (\frac{A - B}{2}) + 1 - 2 {sin}^{2} (\frac{C}{2})$ $2cos(90-C/2)cos((A-B)/2)+1-2sin^2(C/2)$

= $2 sin (\frac{C}{2}) cos (\frac{A - B}{2}) + 1 - 2 {sin}^{2} (\frac{C}{2})$ $2sin(C/2)cos((A-B)/2)+1-2sin^2(C/2)$

= $2 sin (\frac{C}{2}) (cos (\frac{A - B}{2}) - sin (\frac{C}{2})) + 1$ $2sin(C/2)(cos((A-B)/2)-sin(C/2))+1$

= $2 sin (\frac{C}{2}) (cos (\frac{A - B}{2}) - cos (\frac{A + B}{2})) + 1$ $2sin(C/2)(cos((A-B)/2)-cos((A+B)/2))+1$

= $2 sin (\frac{C}{2}) (2 sin (\frac{A}{2}) sin (\frac{B}{2})) + 1$ $2sin(C/2)(2sin(A/2)sin(B/2))+1$

= $1 + 4 sin (\frac{A}{2}) sin (\frac{B}{2}) sin (\frac{C}{2})$ $1+4sin(A/2)sin(B/2)sin(C/2)$