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

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

Solve acosA-bsinA=cacosA-bsinA=c ?