Prove that cosA+cosB+cosC is always positive in triangle ABC.
Determine the proving of the cosA+cosB+cosC that's always the positive in triangle ABC.
Solve the L.H.S part:
cosA+cosB+cosC=(cosA+cosB)+cosC⇒=2·cos[(A+B)2]·cos[(A-B)2]+cosC⇒=2·cos[(π2)–(C2)]·cos[(A-B)2]+cosC⇒=2·sin(C2)·cos[(A-B)2]+1–2·sin²(C2)⇒=1+2sin(C2)·cos[(A-B)2]–sin(C2)⇒=1+2sin(C2)·cos[(A-B)2]–sin[(π2)–((A+B)2)]⇒=1+2sin(C2)·cos[(A-B)2]–cos[(A+B)2]⇒=1+2sin(C2)·2sin(A2)·sin(B2)...............................(ii)⇒=1+4sin(A2)sin(B2)sin(C2)
Hence, the given expression is positive.