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

#cosA+cosB+cosC=1+4sin(A/2)sin(B/2)sin(C/2)#

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

#A+B+C=180# and #A+B=180-C# or #(A+B)/2=90-C/2#

Hence #cosA+cosB+cosC#

= #2cos((A+B)/2)cos((A-B)/2)+1-2sin^2(C/2)-1#

= #2cos(90-C/2)cos((A-B)/2)+1-2sin^2(C/2)#

= #2sin(C/2)cos((A-B)/2)+1-2sin^2(C/2)#

= #2sin(C/2)(cos((A-B)/2)-sin(C/2))+1#

= #2sin(C/2)(cos((A-B)/2)-cos((A+B)/2))+1#

= #2sin(C/2)(2sin(A/2)sin(B/2))+1#

= #1+4sin(A/2)sin(B/2)sin(C/2)#