Question
Question: Define \(\sinh y\)and \(\cosh y\)in terms of exponential functions and show that \(2y = \ln \left...
Define sinhyand coshyin terms of exponential functions and show that
2y = \ln \left\\{ {\dfrac{{\cosh y + \sinh y}}{{\cosh y - \sinh y}}} \right\\}
By putting tanhy=31,deduce that
tanh−1(31)=21ln2
Solution
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→R is defined by sinhy=2[ey−e−y]. Similarly, the hyperbolic cosine function is a function f:R→R is defined by coshy=2[ey+e−y].
Complete step by step solution: The hyperbolic sine function is a function f:R→Ris defined by sinhy=2[ey−e−y]. The hyperbolic cosine function is a function f:R→Ris defined by coshy=2[ey+e−y].
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(31)=21ln2by putting tanhy=31,
We know that the hyperbolic cosine function is a function f:R→Ris defined by
tanhy=ey+e−yey−e−y
Now, taking tanhy=31,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 sinhy and coshy . The expression for tanhy=ey+e−yey−e−y . Similarly we have the expressions for cothy=ey−e−yey+e−y , sechy=ey+e−y2 and cschy=ey−e−y2 .