Question

Arrange the hyperbolic functions in ascending order $A = \sinh 0$ , $B = \operatorname{Cosh} 0$ and $C = \operatorname{Sech} 0$
A. $A,B,C$
B. $A,C,B$
C. $B,C,A$
D. $B,A,C$

Answer 1

Hint : The hyperbolic functions have to be written in terms of exponential functions and the value is to be compared at $x = {0^o}$

Complete step-by-step answer :
The first function is $A = \sinh 0$
The formula for $\sinh x$ in terms of exponential function is,
$\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2}$
Substituting $x = 0$ in equation (1),
$\sinh 0 = \dfrac{{{e^0} - {e^{ - 0}}}}{2} \\\ \sinh 0 = \dfrac{{1 - 1}}{2} \\\ \sinh 0 = 0 \\\$
The value of $A = 0 \cdots \left( 1 \right)$

The second function is $B = \cosh 0$
The formula for in terms of exponential function is,
$\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}$
Substituting in equation (2),
$\cosh 0 = \dfrac{{{e^0} + {e^{ - 0}}}}{2} \\\ \cosh 0 = \dfrac{{1 + 1}}{2} \\\ \cosh 0 = 1 \\\$

The value of $B = 1 \cdots \left( 2 \right)$

The third function is $C = \operatorname{sech} 0$
The formula for $\operatorname{sech} x$ in terms of exponential function is,
$\operatorname{sech} x = \dfrac{1}{{\cosh x}} \\\ \operatorname{sech} x = \dfrac{2}{{{e^x} + {e^{ - x}}}} \\\$

Substituting $x = 0$ in equation (3),
$\operatorname{sech} 0 = \dfrac{2}{{{e^0} + {e^{ - 0}}}} \\\ \operatorname{sech} 0 = \dfrac{2}{2} \\\ \operatorname{sech} 0 = 1 \\\$
The value of $C = 1 \cdots \left( 3 \right)$
From equation (1), (2) and (3), the correct increasing order of the hyperbolic functions is
$\sinh 0,\cosh 0 = \operatorname{sech} 0$ Or $A,B = C$
But this option is not given.
Hence, none of the options is correct.

Note : Hyperbolic functions are very similar to the trigonometric functions but they are expressed in terms of exponential functions. The most common hyperbolic functions are $\sinh x$ and $\cosh x$ .
The formula for in terms of exponential function is , $\cosh x$
$\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}$
This function satisfies the $\cosh 0 = 1$ and $\cosh x = \cosh \left( { - x} \right)$ .
The graph of the $\cosh x$ is always above the graph of $\dfrac{{{e^x}}}{2}$ and $\dfrac{{{e^{ - x} \sinh x }}}{2}$
The formula for in terms of exponential function is ,
$\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2}$
This function satisfies the $\sinh 0 = 0$ and $\sinh \left( { - x} \right) = - \sinh \left( x \right)$ .
The graph of the $\sinh x$ is always between the graphs of $\dfrac{{{e^x}}}{2}$ and $\dfrac{{{e^{ - x}}}}{2}$ .
For the large values of $x$ the graph of and $\cosh x$ are closer to each other .

The value of $\tanh x$ can be calculated by dividing the $\sinh x$ by $\cosh x$ as,
$\tanh x = \dfrac{{\sinh x}}{{\cosh x}} \\\ \tanh x = \dfrac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}} \\\$
The factor of $2$ got cancelled in the numerator and denominator.