If angle A+angle B +angle C = π (180 degrees) ,π-A=B+C
then sinA=sin(π-A)=sin(B+C)
but A might not equal to B+C, right?
Mmm
π−A=B+C→A=π−(B+C)orA=B+C
But I can add 2pi*n to these answers where n is an integer
A=π−(B+C)+2nπ=(2n+1)π−(B+C)orA=B+C+2nπ
Mmm
A=(2n+1)π−(B+C)orA=2nπ+(B+C)$Ithinkthegeneralformulaforthiswouldbe$A=(−1)n(B+C)+nπ
I think that is right but I really need another mathematician to check my working. :))
A+B+C = Pi
B + C = Pi - A
cos(B +C)= cos (Pi -A)
cos (B+C) = -(cos A)
-(cos A) = cos(-A)
so that,
cos(B +C) = cos(-A)
now we cut cos
B+C = -A//
Mmm
π−A=B+C→A=π−(B+C)orA=B+C
But I can add 2pi*n to these answers where n is an integer
A=π−(B+C)+2nπ=(2n+1)π−(B+C)orA=B+C+2nπ
Mmm
A=(2n+1)π−(B+C)orA=2nπ+(B+C)$Ithinkthegeneralformulaforthiswouldbe$A=(−1)n(B+C)+nπ
I think that is right but I really need another mathematician to check my working. :))