Solve-simultaneous-sinh-x-cosh-y-5-sinh-2-x-cosh-2-y-13-

Question Number 178356 by Spillover last updated on 15/Oct/22

Solve simultaneous

\sinh x + \cosh y = 5

\sinh^{2} x + \cosh^{2} y = 13

Answered by Ar Brandon last updated on 16/Oct/22

{\begin{cases} \sinh x + \cosh y = 5 & \dots (i) \\ \sinh^{2} x + \cosh^{2} y = 13 & \dots (i i) \end{cases}

(i i) \Rightarrow 5^{2} - 2 \sinh x \cosh y = 13

\Rightarrow 12 = 2 \sinh x \cosh y

\Rightarrow \sinh x = \frac{6}{\cosh y} \dots (i i i)

(i i i) in (i) \Rightarrow

\frac{6}{\cosh y} + \cosh y = 5 \Rightarrow \cosh^{2} y - 5 \cosh y + 6 = 0

\Rightarrow (\cosh y - 3) (\cosh y - 2) = 0 \Rightarrow \cosh y = 3 or \cosh y = 2

(a) \cosh y = 3

\Rightarrow \frac{1}{2} (e^{y} + e^{- y}) = 3 \Rightarrow e^{2 y} - 6 e^{y} + 1 = 0

\Rightarrow e^{y} = \frac{6 \pm \sqrt{36 - 4}}{2} = 3 \pm 2 \sqrt{2} \Rightarrow y = \ln (3 \pm 2 \sqrt{2})

\Rightarrow \sinh x = 2 \Rightarrow e^{2 x} - 4 e^{x} - 1 = 0 \Rightarrow e^{x} = 2 + \sqrt{5} \Rightarrow x = \ln (2 + \sqrt{5})

(b) \cosh y = 2

\Rightarrow e^{2 y} - 4 e^{y} + 1 = 0 \Rightarrow e^{y} = 2 \pm \sqrt{3} \Rightarrow y = \ln (2 \pm \sqrt{3})

\Rightarrow \sinh x = 3 \Rightarrow e^{2 x} - 6 e^{x} - 1 = 0 \Rightarrow e^{x} = 3 + \sqrt{10} \Rightarrow x = \ln (3 + \sqrt{10})

S = {(x, y) ∣ (x = \ln (2 + \sqrt{5}) \land y = \ln (3 \pm 2 \sqrt{2})) \lor (x = \ln (3 + \sqrt{10}) \land y = \ln (2 \pm \sqrt{3}))}

Commented by Spillover last updated on 16/Oct/22

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