Express the value of the inverse hyperbolic function in terms of natural logarithms:
sech-1(12/13)
??? not sure how to go about this...
+26388 Express the value of the inverse hyperbolic function in terms of natural logarithms:
sech-1(12/13)
see: http://mathworld.wolfram.com/InverseHyperbolicSecant.html
\(\begin{array}{rcll} sech^{-1}{ \left( \frac{12}{13} \right)} &=& \ln{ \left( \dfrac{1+\sqrt{1-(\frac{12}{13})^2}} { \frac{12}{13} } \right)} \\\\ &=& \ln{ \left( \dfrac{1+\sqrt{1-0.92307692308^2}} { 0.92307692308 } \right)} \\\\ &=& \ln{ \left( \dfrac{1+0.38461538462} { 0.92307692308 } \right)} \\\\ &=& \ln{ \left( \dfrac{1.38461538462} { 0.92307692308 } \right)} \\\\ &=& \ln{ ( 1.5 ) } \\\\ \mathbf{ sech^{-1}{ \left( \frac{12}{13} \right)} } & \mathbf{=} & \mathbf{0.40546510811} \end{array}\)
+26388 Express the value of the inverse hyperbolic function in terms of natural logarithms:
sech-1(12/13)
see: http://mathworld.wolfram.com/InverseHyperbolicSecant.html
\(\begin{array}{rcll} sech^{-1}{ \left( \frac{12}{13} \right)} &=& \ln{ \left( \dfrac{1+\sqrt{1-(\frac{12}{13})^2}} { \frac{12}{13} } \right)} \\\\ &=& \ln{ \left( \dfrac{1+\sqrt{1-0.92307692308^2}} { 0.92307692308 } \right)} \\\\ &=& \ln{ \left( \dfrac{1+0.38461538462} { 0.92307692308 } \right)} \\\\ &=& \ln{ \left( \dfrac{1.38461538462} { 0.92307692308 } \right)} \\\\ &=& \ln{ ( 1.5 ) } \\\\ \mathbf{ sech^{-1}{ \left( \frac{12}{13} \right)} } & \mathbf{=} & \mathbf{0.40546510811} \end{array}\)
