If the relation between two inverse hyperbolic functions is given as ${{\sinh }^{-1}}\left( 2\alpha \right)={{\cosh }^{-1}}\left( \beta \right)$, then find the relation between $\alpha $ and $\beta $?
(a) ${{\alpha }^{2}}+{{\beta }^{2}}={{\alpha }^{4}}$,
(b) ${{\beta }^{2}}-4{{\alpha }^{2}}=1$,
(c) ${{\alpha }^{2}}+{{\beta }^{2}}={{\beta }^{2}}$,
(d) ${{\alpha }^{2}}={{\beta }^{2}}$.
Answer
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Hint: We start solving the problem by recalling the definitions of ${{\sinh }^{-1}}\left( x \right)$ and ${{\cosh }^{-1}}\left( x \right)$. We apply these definitions to the given relation ${{\sinh }^{-1}}\left( 2\alpha \right)={{\cosh }^{-1}}\left( \beta \right)$. We then bring the terms with square root on one side and the terms with out square root on other side. We square them on both sides and make necessary calculations and repeat the previous step to get the desired solution.
Complete step by step answer:
According to the problem, we are given the relation between two inverse hyperbolic functions as ${{\sinh }^{-1}}\left( 2\alpha \right)={{\cosh }^{-1}}\left( \beta \right)$. We need to find the relation between $\alpha $ and $\beta $.
Let us recall the definitions of ${{\sinh }^{-1}}\left( x \right)$ and ${{\cosh }^{-1}}\left( x \right)$.
We know that ${{\sinh }^{-1}}\left( x \right)={{\log }_{e}}\left( x+\sqrt{{{x}^{2}}+1} \right)$ and ${{\cosh }^{-1}}\left( x \right)={{\log }_{e}}\left( x+\sqrt{{{x}^{2}}-1} \right)$. We use this in the given relation ${{\sinh }^{-1}}\left( 2\alpha \right)={{\cosh }^{-1}}\left( \beta \right)$.
$\Rightarrow {{\log }_{e}}\left( 2\alpha +\sqrt{{{\left( 2\alpha \right)}^{2}}+1} \right)={{\log }_{e}}\left( \beta +\sqrt{{{\beta }^{2}}-1} \right)$.
$\Rightarrow {{\log }_{e}}\left( 2\alpha +\sqrt{4{{\alpha }^{2}}+1} \right)={{\log }_{e}}\left( \beta +\sqrt{{{\beta }^{2}}-1} \right)$.
$\Rightarrow {{\log }_{e}}\left( 2\alpha +\sqrt{4{{\alpha }^{2}}+1} \right)-{{\log }_{e}}\left( \beta +\sqrt{{{\beta }^{2}}-1} \right)=0$.
We know that ${{\log }_{a}}x-{{\log }_{a}}y={{\log }_{a}}\dfrac{x}{y}$. For a positive value of a, x and y.
\[\Rightarrow {{\log }_{e}}\left( \dfrac{\left( 2\alpha +\sqrt{4{{\alpha }^{2}}+1} \right)}{\left( \beta +\sqrt{{{\beta }^{2}}-1} \right)} \right)=0\].
We know that if ${{\log }_{a}}x=0$, then $x=1$.
So, we get \[\dfrac{2\alpha +\sqrt{4{{\alpha }^{2}}+1}}{\beta +\sqrt{{{\beta }^{2}}-1}}=1\].
$\Rightarrow 2\alpha +\sqrt{4{{\alpha }^{2}}+1}=\beta +\sqrt{{{\beta }^{2}}-1}$.
$\Rightarrow 2\alpha -\beta =\sqrt{{{\beta }^{2}}-1}-\sqrt{4{{\alpha }^{2}}+1}$.
Now, let us square on both sides.
$\Rightarrow {{\left( 2\alpha -\beta \right)}^{2}}={{\left( \sqrt{{{\beta }^{2}}-1}-\sqrt{4{{\alpha }^{2}}+1} \right)}^{2}}$.
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$.
$\Rightarrow {{\left( 2\alpha \right)}^{2}}+{{\left( \beta \right)}^{2}}-2\left( 2\alpha \right)\left( \beta \right)={{\left( \sqrt{{{\beta }^{2}}-1} \right)}^{2}}+{{\left( \sqrt{4{{\alpha }^{2}}+1} \right)}^{2}}-2\left( \sqrt{{{\beta }^{2}}-1} \right)\left( \sqrt{4{{\alpha }^{2}}+1} \right)$.
$\Rightarrow 4{{\alpha }^{2}}+{{\beta }^{2}}-4\alpha \beta ={{\beta }^{2}}-1+4{{\alpha }^{2}}+1-2\left( \sqrt{\left( {{\beta }^{2}}-1 \right)\left( 4{{\alpha }^{2}}+1 \right)} \right)$.
$\Rightarrow 4{{\alpha }^{2}}+{{\beta }^{2}}-{{\beta }^{2}}-4{{\alpha }^{2}}-4\alpha \beta =-2\left( \sqrt{4{{\alpha }^{2}}{{\beta }^{2}}-4{{\alpha }^{2}}+{{\beta }^{2}}-1} \right)$.
$\Rightarrow -4\alpha \beta =-2\left( \sqrt{4{{\alpha }^{2}}{{\beta }^{2}}-4{{\alpha }^{2}}+{{\beta }^{2}}-1} \right)$.
$\Rightarrow 2\alpha \beta =\sqrt{4{{\alpha }^{2}}{{\beta }^{2}}-4{{\alpha }^{2}}+{{\beta }^{2}}-1}$.
Let us again square on both sides.
$\Rightarrow {{\left( 2\alpha \beta \right)}^{2}}={{\left( \sqrt{4{{\alpha }^{2}}{{\beta }^{2}}-4{{\alpha }^{2}}+{{\beta }^{2}}-1} \right)}^{2}}$.
$\Rightarrow 4{{\alpha }^{2}}{{\beta }^{2}}=4{{\alpha }^{2}}{{\beta }^{2}}-4{{\alpha }^{2}}+{{\beta }^{2}}-1$.
$\Rightarrow {{\beta }^{2}}-4{{\alpha }^{2}}=4{{\alpha }^{2}}{{\beta }^{2}}-4{{\alpha }^{2}}{{\beta }^{2}}+1$.
$\Rightarrow {{\beta }^{2}}-4{{\alpha }^{2}}=1$.
So, we have found the relation between the $\alpha $ and $\beta $ as ${{\beta }^{2}}-4{{\alpha }^{2}}=1$.
∴ The correct option for the given problem is (b).
Note:
Alternatively, we can solve this problem as shown below,
We have given ${{\sinh }^{-1}}\left( 2\alpha \right)={{\cosh }^{-1}}\left( \beta \right)$.
\[\Rightarrow 2\alpha =\sinh \left( {{\cosh }^{-1}}\left( \beta \right) \right)\].
Let us assume ${{\cosh }^{-1}}\left( \beta \right)=x$. So, we get $\cosh x=\beta $.
\[\Rightarrow 2\alpha =\sinh \left( x \right)\] ---(1).
We know that ${{\cosh }^{2}}x-{{\sinh }^{2}}x=1$.
$\Rightarrow {{\beta }^{2}}-{{\sinh }^{2}}x=1$.
$\Rightarrow {{\beta }^{2}}-1={{\sinh }^{2}}x$.
$\Rightarrow \sinh x=\sqrt{{{\beta }^{2}}-1}$. Let us substitute this in equation (1).
\[\Rightarrow 2\alpha =\sqrt{{{\beta }^{2}}-1}\].
Let us square on both sides.
\[\Rightarrow {{\left( 2\alpha \right)}^{2}}={{\left( \sqrt{{{\beta }^{2}}-1} \right)}^{2}}\].
\[\Rightarrow 4{{\alpha }^{2}}={{\beta }^{2}}-1\].
$\Rightarrow {{\beta }^{2}}-4{{\alpha }^{2}}=1$.
Complete step by step answer:
According to the problem, we are given the relation between two inverse hyperbolic functions as ${{\sinh }^{-1}}\left( 2\alpha \right)={{\cosh }^{-1}}\left( \beta \right)$. We need to find the relation between $\alpha $ and $\beta $.
Let us recall the definitions of ${{\sinh }^{-1}}\left( x \right)$ and ${{\cosh }^{-1}}\left( x \right)$.
We know that ${{\sinh }^{-1}}\left( x \right)={{\log }_{e}}\left( x+\sqrt{{{x}^{2}}+1} \right)$ and ${{\cosh }^{-1}}\left( x \right)={{\log }_{e}}\left( x+\sqrt{{{x}^{2}}-1} \right)$. We use this in the given relation ${{\sinh }^{-1}}\left( 2\alpha \right)={{\cosh }^{-1}}\left( \beta \right)$.
$\Rightarrow {{\log }_{e}}\left( 2\alpha +\sqrt{{{\left( 2\alpha \right)}^{2}}+1} \right)={{\log }_{e}}\left( \beta +\sqrt{{{\beta }^{2}}-1} \right)$.
$\Rightarrow {{\log }_{e}}\left( 2\alpha +\sqrt{4{{\alpha }^{2}}+1} \right)={{\log }_{e}}\left( \beta +\sqrt{{{\beta }^{2}}-1} \right)$.
$\Rightarrow {{\log }_{e}}\left( 2\alpha +\sqrt{4{{\alpha }^{2}}+1} \right)-{{\log }_{e}}\left( \beta +\sqrt{{{\beta }^{2}}-1} \right)=0$.
We know that ${{\log }_{a}}x-{{\log }_{a}}y={{\log }_{a}}\dfrac{x}{y}$. For a positive value of a, x and y.
\[\Rightarrow {{\log }_{e}}\left( \dfrac{\left( 2\alpha +\sqrt{4{{\alpha }^{2}}+1} \right)}{\left( \beta +\sqrt{{{\beta }^{2}}-1} \right)} \right)=0\].
We know that if ${{\log }_{a}}x=0$, then $x=1$.
So, we get \[\dfrac{2\alpha +\sqrt{4{{\alpha }^{2}}+1}}{\beta +\sqrt{{{\beta }^{2}}-1}}=1\].
$\Rightarrow 2\alpha +\sqrt{4{{\alpha }^{2}}+1}=\beta +\sqrt{{{\beta }^{2}}-1}$.
$\Rightarrow 2\alpha -\beta =\sqrt{{{\beta }^{2}}-1}-\sqrt{4{{\alpha }^{2}}+1}$.
Now, let us square on both sides.
$\Rightarrow {{\left( 2\alpha -\beta \right)}^{2}}={{\left( \sqrt{{{\beta }^{2}}-1}-\sqrt{4{{\alpha }^{2}}+1} \right)}^{2}}$.
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$.
$\Rightarrow {{\left( 2\alpha \right)}^{2}}+{{\left( \beta \right)}^{2}}-2\left( 2\alpha \right)\left( \beta \right)={{\left( \sqrt{{{\beta }^{2}}-1} \right)}^{2}}+{{\left( \sqrt{4{{\alpha }^{2}}+1} \right)}^{2}}-2\left( \sqrt{{{\beta }^{2}}-1} \right)\left( \sqrt{4{{\alpha }^{2}}+1} \right)$.
$\Rightarrow 4{{\alpha }^{2}}+{{\beta }^{2}}-4\alpha \beta ={{\beta }^{2}}-1+4{{\alpha }^{2}}+1-2\left( \sqrt{\left( {{\beta }^{2}}-1 \right)\left( 4{{\alpha }^{2}}+1 \right)} \right)$.
$\Rightarrow 4{{\alpha }^{2}}+{{\beta }^{2}}-{{\beta }^{2}}-4{{\alpha }^{2}}-4\alpha \beta =-2\left( \sqrt{4{{\alpha }^{2}}{{\beta }^{2}}-4{{\alpha }^{2}}+{{\beta }^{2}}-1} \right)$.
$\Rightarrow -4\alpha \beta =-2\left( \sqrt{4{{\alpha }^{2}}{{\beta }^{2}}-4{{\alpha }^{2}}+{{\beta }^{2}}-1} \right)$.
$\Rightarrow 2\alpha \beta =\sqrt{4{{\alpha }^{2}}{{\beta }^{2}}-4{{\alpha }^{2}}+{{\beta }^{2}}-1}$.
Let us again square on both sides.
$\Rightarrow {{\left( 2\alpha \beta \right)}^{2}}={{\left( \sqrt{4{{\alpha }^{2}}{{\beta }^{2}}-4{{\alpha }^{2}}+{{\beta }^{2}}-1} \right)}^{2}}$.
$\Rightarrow 4{{\alpha }^{2}}{{\beta }^{2}}=4{{\alpha }^{2}}{{\beta }^{2}}-4{{\alpha }^{2}}+{{\beta }^{2}}-1$.
$\Rightarrow {{\beta }^{2}}-4{{\alpha }^{2}}=4{{\alpha }^{2}}{{\beta }^{2}}-4{{\alpha }^{2}}{{\beta }^{2}}+1$.
$\Rightarrow {{\beta }^{2}}-4{{\alpha }^{2}}=1$.
So, we have found the relation between the $\alpha $ and $\beta $ as ${{\beta }^{2}}-4{{\alpha }^{2}}=1$.
∴ The correct option for the given problem is (b).
Note:
Alternatively, we can solve this problem as shown below,
We have given ${{\sinh }^{-1}}\left( 2\alpha \right)={{\cosh }^{-1}}\left( \beta \right)$.
\[\Rightarrow 2\alpha =\sinh \left( {{\cosh }^{-1}}\left( \beta \right) \right)\].
Let us assume ${{\cosh }^{-1}}\left( \beta \right)=x$. So, we get $\cosh x=\beta $.
\[\Rightarrow 2\alpha =\sinh \left( x \right)\] ---(1).
We know that ${{\cosh }^{2}}x-{{\sinh }^{2}}x=1$.
$\Rightarrow {{\beta }^{2}}-{{\sinh }^{2}}x=1$.
$\Rightarrow {{\beta }^{2}}-1={{\sinh }^{2}}x$.
$\Rightarrow \sinh x=\sqrt{{{\beta }^{2}}-1}$. Let us substitute this in equation (1).
\[\Rightarrow 2\alpha =\sqrt{{{\beta }^{2}}-1}\].
Let us square on both sides.
\[\Rightarrow {{\left( 2\alpha \right)}^{2}}={{\left( \sqrt{{{\beta }^{2}}-1} \right)}^{2}}\].
\[\Rightarrow 4{{\alpha }^{2}}={{\beta }^{2}}-1\].
$\Rightarrow {{\beta }^{2}}-4{{\alpha }^{2}}=1$.
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