if A+B+C=180 degree,show that determinant whose r1=[sin2 A sinA cos A - Maths - Determinants

Question

if A+B+C=180 degree,show that determinant whose r1=[sin2
A sinA cos A cos2 A], r2=[sin2 B sinB cos B
cos​2 B] and r3=[sin2C sinC cosC
cos2C] = -sin(A-B) sin(B-C) sin(C-A)

Sanjay Kumar Manjhi · Accepted Answer

Dear Students,If A+B+C=π, show that;sin2AsinAcosAcos2Asin2BsinBcosBcos2Bsin2CsinCcosCcos2C=-sin(A-B)sin(B-C)sin(C-A)Now, C1→C1+C3 and multiply C2 and C3 by 2;=14sin2A+cos2A2sinAcosA2cos2Asin2B+cos2B2sinBcosB2cos2B sin2C+cos2C2sinCcosC2cos2C         (Here 14 is used outside the determinant to counter the 2 multiplied in C2 and C3)=141sin2A1+cos2A1sin2B1+cos2B1sin2C1+cos2C            (Because, 2cos2x=1+cos2x)Now, R2→R2-R1 and R3→R3-R1=141sin2A1+cos2A0sin2B-sin2Acos2B-cos2A0sin2C-sin2Acos2C-cos2A=141sin2A1+cos2A02cos(A+B)sin(B-A)2sin(A+B)sin(A-B)02cos(A+C)sin(C-A)2sin(A+C)sin(A-C)         {Because, sinX-sinY=2sinX-Y2cosX+Y2 and cosX-cosY=2sinY-X2sinX+Y2}take common 2sin(A-B) and 2sin(A-C) from R2 and R3 respectively;=142sin(A-B)×2sin(A-C)1sin2A1+cos2A0-cos(A+B)sin(A+B)0-cos(A+C)sin(A+C)=sin(A-B)×sin(A-C)1sin2A1+cos2A0-cos(π-C)sin(π-C)0-cos(π-B)sin(π-B)             { Because, A+B+C=π}=sin(A-B)×sin(A-C)1sin2A1+cos2A0cosCsinC0cosBsinB=sin(A-B)×sin(A-C)[cosCsinB-sinCcosB]=sin(A-B)×sin(A-C)×sin(B-C)=-sin(A-B)×sin(B-C)×sin(C-A) =RHS, Hence Proved
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if A+B+C=180 degree,show that determinant whose r1=[sin2 A sinA cos A cos2 A], r2=[sin2 B sinB cos B cos​2 B] and r3=[sin2C sinC cosC cos2C] = -sin(A-B) sin(B-C) sin(C-A).

if A+B+C=180 degree,show that determinant whose r1=[sin² A sinA cos A cos² A], r2=[sin² B sinB cos B cos² B] and r3=[sin²C sinC cosC cos²C] = -sin(A-B) sin(B-C) sin(C-A).