Question

Define $\sinh y$and $\cosh y$in terms of exponential functions and show that
$2y = \ln \left\{ {\dfrac{{\cosh y + \sinh y}}{{\cosh y - \sinh y}}} \right\}$
By putting $\tanh y = \dfrac{1}{3},$deduce that
${\tanh ^{ - 1}}\left( {\dfrac{1}{3}} \right) = \dfrac{1}{2}\ln 2$

Answer 1

Hint: Now, in this question we are given the hyperbolic functions of sin, cos and tan.
Now the hyperbolic sine function is a function $f: R \to R$ is defined by $\sinh y = \dfrac{{\left[ {{e^y} - {e^{ - y}}} \right]}}{2}$. Similarly, the hyperbolic cosine function is a function $f:R \to R$ is defined by $\cosh y = \dfrac{{\left[ {{e^y} + {e^{ - y}}} \right]}}{2}$.

Complete step by step solution: The hyperbolic sine function is a function $f:R \to R$is defined by $\sinh y = \dfrac{{\left[ {{e^y} - {e^{ - y}}} \right]}}{2}$. The hyperbolic cosine function is a function $f:R \to R$is defined by $\cosh y = \dfrac{{\left[ {{e^y} + {e^{ - y}}} \right]}}{2}$.
Now, we need to show that $2y = \ln \left\{ {\dfrac{{\cosh y + \sinh y}}{{\cosh y - \sinh y}}} \right\}$
Now, starting with the RHS, simplifying it:
$ RHS: \\
  \ln \left\{ {\dfrac{{\cosh y + \sinh y}}{{\cosh y - \sinh y}}} \right\} \\
= \ln \left( {\dfrac{{\dfrac{{{e^y} + {e^{ - y}}}}{2} + \dfrac{{{e^y} - {e^{ - y}}}}{2}}}{{\dfrac{{{e^y} + {e^{ - y}}}}{2} - \dfrac{{{e^y} - {e^{ - y}}}}{2}}}} \right) \\
= \ln \left( {\dfrac{{\dfrac{{2{e^y}}}{2}}}{{\dfrac{{2{e^{ - y}}}}{2}}}} \right) \\
= \ln \left( {\dfrac{{{e^y}}}{{{e^{ - y}}}}} \right) \\
= \ln \left( {{e^y}.{e^y}} \right) \\
= \ln \left( {{e^{2y}}} \right) \\
= 2y \\
= LHS \\ $
Therefore we have proved that
 $2y = \ln \left\{ {\dfrac{{\cosh y + \sinh y}}{{\cosh y - \sinh y}}} \right\}$
Next we need to deduce that
${\tanh ^{ - 1}}\left( {\dfrac{1}{3}} \right) = \dfrac{1}{2}\ln 2$by putting $\tanh y = \dfrac{1}{3},$
We know that the hyperbolic cosine function is a function $f:R \to R$is defined by
$\tanh y = \dfrac{{{e^y} - {e^{ - y}}}}{{{e^y} + {e^{ - y}}}}$
Now, taking $\tanh y = \dfrac{1}{3},$and putting in the formula we will get:
$ \dfrac{1}{3} = \dfrac{{{e^y} - {e^{ - y}}}}{{{e^y} + {e^{ - y}}}} \\
\Rightarrow \dfrac{1}{3} = \dfrac{{{e^y} - \dfrac{1}{{{e^y}}}}}{{{e^y} + \dfrac{1}{{{e^y}}}}} \\
\Rightarrow \dfrac{1}{3} = \dfrac{{\dfrac{{{e^{2y}} - 1}}{{{e^y}}}}}{{\dfrac{{{e^2} + 1}}{{{e^2}}}}} \\
\Rightarrow \dfrac{1}{3} = \dfrac{{{e^{2y}} - 1}}{{{e^{2y}} + 1}} \\
\Rightarrow {e^{2y}} + 1 = 3({e^{2y}} - 1) \\
 \Rightarrow {e^{2y}} + 1 = 3{e^{2y}} - 3 \\
 \Rightarrow 4 = 2{e^{2y}} \\
 \Rightarrow {e^{2y}} = 2 \\
 \Rightarrow 2y = \ln 2 \\
\Rightarrow y = \dfrac{1}{2}\ln 2 \\ $
Therefore, we have deduced the relation successfully.

Note: We are working with hyperbolic functions, so the general trigonometric rules or formulae will not apply. The hyperbolic trigonometric functions have exponential values. Here we have used $\sinh y$ and $\cosh y$ . The expression for $\tanh y = \dfrac{{{e^y} - {e^{ - y}}}}{{{e^y} + {e^{ - y}}}}$ . Similarly we have the expressions for $\coth y = \dfrac{{{e^y} + {e^{ - y}}}}{{{e^y} - {e^{ - y}}}}$ , $\operatorname{sech} y = \dfrac{2}{{{e^y} + {e^{ - y}}}}$ and $\operatorname{csch} y = \dfrac{2}{{{e^y} - {e^{ - y}}}}$ .