# Solve acosA-bsinA=c ?

Feb 3, 2017

given

$a \cos A - b \sin A = c$

Squaring both sides we get

$\implies {\left(a \cos A - b \sin A\right)}^{2} = {c}^{2}$

$\implies {a}^{2} {\cos}^{2} A + {b}^{2} {\sin}^{2} A - 2 a b \sin A \cos A = {c}^{2}$

$\implies {a}^{2} \left(1 - {\sin}^{2} A\right) + {b}^{2} \left(1 - {\cos}^{2} A\right) - 2 a b \sin A \cos A = {c}^{2}$

$\implies {a}^{2} - {a}^{2} {\sin}^{2} A + {b}^{2} - {b}^{2} {\cos}^{2} A - 2 a b \sin A \cos A = {c}^{2}$

$\implies {a}^{2} {\sin}^{2} A + {b}^{2} {\cos}^{2} A + 2 a b \sin A \cos A = {a}^{2} + {b}^{2} - {c}^{2}$

$\implies {\left(a \sin A + b \cos A\right)}^{2} = {a}^{2} + {b}^{2} - {c}^{2}$

$\implies a \sin A + b \cos A = \pm \sqrt{{a}^{2} + {b}^{2} - {c}^{2}}$

Feb 3, 2017

$\pm \sqrt{{a}^{2} + {b}^{2} - {c}^{2}}$

#### Explanation:

$\left\{\begin{matrix}a \cos A - b \sin A = c \\ i a \sin A + i b \cos A = i x\end{matrix}\right.$

$a \left(\cos A + i \sin A\right) + i b \left(\cos A + i \sin A\right) = c + i x$

or

$\left(a + i b\right) {e}^{i A} = c + i x$ Now, multiplying by the conjugate we have

$\left(a + i b\right) \left(a - i b\right) {e}^{i A} {e}^{- i A} = \left(c + i x\right) \left(c - i x\right)$

so

${a}^{2} + {b}^{2} = {c}^{2} + {x}^{2}$ and finally

$x = \pm \sqrt{{a}^{2} + {b}^{2} - {c}^{2}}$