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sinh^-1 (1)

Guest Sep 14, 2017
 #1
avatar+7517 
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sinh^-1 (1)

 

\(sinh^{-1}(1°)_{deg}=asinh(1°)_{deg}\color{blue}=50.498986710526\\ sinh^{-1}(1)_{rad}=asinh(1)_{rad}\color{blue}=0.88137358702\)

 

laugh  !

asinus  Sep 14, 2017
edited by asinus  Sep 14, 2017
 #2
avatar+20151 
0

hyperbolic functions

\(sinh^{-1} (1)\)

 

Formula:

\(\begin{array}{|rcll|} \hline sinh(z) &=& \frac{e^z-e^{-z}}{2} \\ z &=& sinh^{-1} (\frac{e^z-e^{-z}}{2} ) \\ \hline \end{array}\)

 

\(z=\ ?\)

\(\begin{array}{|rcll|} \hline \frac{e^z-e^{-z}}{2} &=& 1 \\ e^z-e^{-z} &=& 2 \\ e^z-\frac{1}{e^{z}} &=& 2 \quad & | \quad \text{substitute } e^z = x \\ x-\frac{1}{x} &=& 2 \quad & | \quad \cdot x \\ x^2-1 &=& 2x \\ x^2-2x-1 &=& 0 \\\\ x &=& \frac{2\pm \sqrt{4-4\cdot(-1)} }{2} \\ &=& \frac{2\pm \sqrt{8} }{2} \\ &=& \frac{2\pm \sqrt{2\cdot 4 } }{2} \\ &=& \frac{2\pm 2\sqrt{2} }{2} \\ \mathbf{x} &\mathbf{=}& \mathbf{1\pm \sqrt{2}} \quad & | \quad x = e^z \\ e^z & = & 1\pm \sqrt{2} \quad & | \quad \ln() \\ z & =& \ln(1\pm \sqrt{2}) \\\\ z_1 &=& \ln(1 + \sqrt{2}) \\ z_2 &=& \ln(1 - \sqrt{2}) \quad & | \quad \text{no solution } \quad 1 - \sqrt{2} < 0! \\\\ \mathbf{sinh^{-1} (1)= z} &\mathbf{=}& \mathbf{\ln(1 + \sqrt{2})} \\ \mathbf{sinh^{-1} (1) } &\mathbf{=}& \mathbf{0.88137358702\ldots} \\ \hline \end{array}\)

 

laugh

heureka  Sep 14, 2017

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