Solve #acosA-bsinA=c# ?

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Answer 1 · 2017-02-03T13:06:44

given

#acosA-bsinA=c#

Squaring both sides we get

#=>(acosA-bsinA)^2=c^2#

#=>a^2cos^2A+b^2sin^2A-2ab sinAcosA=c^2#

#=>a^2(1-sin^2A)+b^2(1-cos^2A)-2ab sinAcosA=c^2#

#=>a^2-a^2sin^2A+b^2-b^2cos^2A-2ab sinAcosA=c^2#

#=>a^2sin^2A+b^2cos^2A+2ab sinAcosA=a^2+b^2-c^2#

#=>(asinA+bcosA)^2=a^2+b^2-c^2#

#=>asinA+bcosA=pmsqrt(a^2+b^2-c^2)#

Answer 2 · 2017-02-03T15:42:12

#pmsqrt(a^2+b^2-c^2)#

#{(acosA-bsinA=c),(iasinA+ibcosA=ix):}#

Adding we have

#a(cosA+isinA)+ib(cosA+isinA)=c+ix#

or

#(a+ib)e^(iA)=c+ix# Now, multiplying by the conjugate we have

#(a+ib)(a-ib)e^(iA)e^(-iA)=(c+ix)(c-ix)#

so

#a^2+b^2=c^2+x^2# and finally

#x=pmsqrt(a^2+b^2-c^2)#