Answers

2014-08-11T21:48:48+05:30
Given A+B+C=180
LHS      SIN²A/2+SIN²B/2+SIN²C/2
            1/2(2SIN²A/2+2SIN²B/2+2SIN²C/2)
            1/2(1-COSA+1-COSB+1-COSC)
            1/2(3-(COSA+COSB+COSC)
            1/2(3- (2COS (A+B/2)COS(A-B/2)+COSC)
            1/2(3- (2COS(90-C/2)COS(A+B/2)+C0SC)
            1/2(3- (2SINC/2COS(A-B/2)+1-2SIN²C/2)
            1/2(3-1-2SINC/2(COS(A+B/2)-SINC/2)
            1/2(2-2SINC/2(COS(A-B/2)-SIN (90-A+B/2)  (SIN90-Ф)= COSФ)
            1/2(2-2SINC/2(COS(A-B/2)-COS(A+B/2)
            1/2(2-2SINC/2(2SINA/2SINB/2)
            1/2(2-4SINA/2SINB/2SINC/2)
            1-2SINA/2SINB/2SINC/2
            RHS       
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