A+B-C=3pi then find sinA+sinB-sinC
1 answer:
The answer: the full question is as follow: if A+B-C=3pi, then find sinA+sinB-sinC first, the main formula of sine and cosine are: sinC = 2sin(C/2)cos(C/2) sinA +sinB = 2sin[(A+B)/2]cos[(A-B)/2] therefore: sinA+sinB-sinC = 2sin[(A+B)/2]cos[(A-B)/2] - 2sin(C/2)cos(C/2) sin[(A+B)/2] = cos(C/2) 2sin(C/2)cos(C/2) = cos[(A+B)/2 and A+B-C=3 pi implies A+B =3 pi + C, so cos[(A+B)/2] = cos [3 pi/2 + C/2] and with the equivalence cos (3Pi/2 + X) = sinX sinA+sinB-sinC = cos(C/2)+ sin(C/2)
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