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Let A, B, C be the angles of a triangle, so that A + B + C = pi. Show that cos(A+B) = -cosC
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subtract C from both side ..u will get
\[\huge A+B=\pi-c\]
now take cosine both side
\[\huge \cos (\pi-C) =-cosC\]
cos(A+B)+cos(C)=0 cos(A+B)+cos(c)=0 2 cos ((A+B+C)/2) cos ( (A+B-C)/2)=0 cos ((A+B+C)/2) = cos(pi/2) =0 done :)
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