# If (acosA - bsinA) = c , prove that (asinA + bcosA) = + or - (a2 + b2 - c2 )1/2

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### 2 Answers

`a*cos(A)-b*sin(A)=c`

`(a*cos(A)-b*sin(A))^2=a^2cos^2(A)-2abcos(A)sin(A)+b^2sin^2(A)=c^2`

`a^2(1-sin^2(A))-2abcos(A)sin(A)+b^2(1-cos^2(A))=c^2`

`a^2-a^2sin^2(A)-2abcos(A)sin(A)+b^2-b^2cos^2(A)=c^2`

`-(a^2sin^2(A)+2abcos(A)sin(A)+b^2cos^2(A))=c^2-a^2-b^2`

`-(asin(A)+bcos(A))^2=c^2-a^2-b^2`

`(asin(A)+bcos(A))^2=a^2+b^2-c^2`

`asin(A)+bcos(A)=+-sqrt(a^2+b^2-c^2)`

squaring on both the sides and taking theta as A,we get-

a2cos2A + b2sin2A - 2ab sinAcosA = c2

hence, 2absinAcosA= a2cos2A + b2sin2A-c2 .............................................(1)

now, asinA+bcosA=(asinA+bcosA)^2*1\2

=(a2sin2A+b2cos2A+2ab sinA cosA)1\2

=(a2sin2A+b2cos2A+ a2cos2A + b2sin2A- c2)1\2

=(a2sin2A + a2cos2A+b2cos2A + b2sin2A- c2)1\2

=[a2(sin2A+cos2A)+b2(sin2A+cos2A)- c2]1\2

as sin2A+cos2A = 1

asinA + bcosA = +_ ( a2 + b2 - c2)1/2