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Question: Define \(\sinh y\)and \(\cosh y\)in terms of exponential functions and show that \(2y = \ln \left...

Define sinhy\sinh yand coshy\cosh yin terms of exponential functions and show that
2y = \ln \left\\{ {\dfrac{{\cosh y + \sinh y}}{{\cosh y - \sinh y}}} \right\\}
By putting tanhy=13,\tanh y = \dfrac{1}{3},deduce that
tanh1(13)=12ln2{\tanh ^{ - 1}}\left( {\dfrac{1}{3}} \right) = \dfrac{1}{2}\ln 2

Explanation

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:RRf: R \to R is defined by sinhy=[eyey]2\sinh y = \dfrac{{\left[ {{e^y} - {e^{ - y}}} \right]}}{2}. Similarly, the hyperbolic cosine function is a function f:RRf:R \to R is defined by coshy=[ey+ey]2\cosh y = \dfrac{{\left[ {{e^y} + {e^{ - y}}} \right]}}{2}.

Complete step by step solution: The hyperbolic sine function is a function f:RRf:R \to Ris defined by sinhy=[eyey]2\sinh y = \dfrac{{\left[ {{e^y} - {e^{ - y}}} \right]}}{2}. The hyperbolic cosine function is a function f:RRf:R \to Ris defined by coshy=[ey+ey]2\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
tanh1(13)=12ln2{\tanh ^{ - 1}}\left( {\dfrac{1}{3}} \right) = \dfrac{1}{2}\ln 2by putting tanhy=13,\tanh y = \dfrac{1}{3},
We know that the hyperbolic cosine function is a function f:RRf:R \to Ris defined by
tanhy=eyeyey+ey\tanh y = \dfrac{{{e^y} - {e^{ - y}}}}{{{e^y} + {e^{ - y}}}}
Now, taking tanhy=13,\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 sinhy\sinh y and coshy\cosh y . The expression for tanhy=eyeyey+ey\tanh y = \dfrac{{{e^y} - {e^{ - y}}}}{{{e^y} + {e^{ - y}}}} . Similarly we have the expressions for cothy=ey+eyeyey\coth y = \dfrac{{{e^y} + {e^{ - y}}}}{{{e^y} - {e^{ - y}}}} , sechy=2ey+ey\operatorname{sech} y = \dfrac{2}{{{e^y} + {e^{ - y}}}} and cschy=2eyey\operatorname{csch} y = \dfrac{2}{{{e^y} - {e^{ - y}}}} .