Question

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

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 $\sinh y={\displaystyle \frac{\left[e^y-e^{-y}\right]}{2}}$. Similarly, the hyperbolic cosine function is a function $f:R\to R$ is defined by $\cosh y={\displaystyle \frac{\left[e^y+e^{-y}\right]}{2}}$.

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