`A+B+C=180^o``, cos 2A+cos2B+cos2C=`

A+B+C=180^(@),cos2A+cos2B+cos2C=

If A+B+C=180^(@) then cos2A+cos2B+cos2C=1-4cos A cos B cos C 2.If A+B+C=0 then cos A+cos B+cos C=-1+4(cos A)/(2)(cos B)/(2)(cos C)/(2)

If A+B+C=180^(@) then cos A+cos B-cosC=

If A+B+C=180^(@) , then (cos2A+cos2B+cos2C+1)/(cos A cos B cos C)

A + B + C = 180 ^ (@) rArr cos2A + cos2B + cos2C =

If A+ B+C= 180^(@) " then " cos 2A+ cos 2B + cos 2C+ 1=

If A + B + C =180^@ , prove that : cos (-A+B +C)+cos(A-B+C)+ cos (A +B-C)=1+ 4cos A cos B cos C .

If A+B+C=180^(@) then cos A+cos B-cos C=

If A+B+C=270^@ , then cos2A+cos2B+cos2C+4sinAsin B sinC=