Question

Find the value of \[{\sinh ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \]
( 1) \[{\tanh ^{ - 1}}\sqrt 5 \]
( 2) \[{\tanh ^{ - 1}}\left( {\dfrac{1}{{\sqrt 5 }}} \right)\]
( 3) \[{\tanh ^{ - 1}}\sqrt 3 \]
( 4) \[{\tanh ^{ - 1}}\left( {\dfrac{2}{{\sqrt 5 }}} \right)\]

Answer 1

Hint: First we have to keep in mind that it is inverse \[\sin \]hyperbolic or \[{\sinh ^{ - 1}}\] function ; not inverse \[\sin \]function . Then we should obtain the value of \[{\sinh ^{ - 1}}\left( {\dfrac{1}{2}} \right)\] using formula . After finding that value we have to equate that with the value of \[{\tanh ^{ - 1}}x\] . We have to use the formula of \[{\tanh ^{ - 1}}x\] and find the value of \[x\]. For simplification and calculation we have to use some basic logarithmic and fraction formulas .
FORMULA USED
 \[{\sinh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} + 1} } \right)\]
\[{\tanh ^{ - 1}}x = \dfrac{1}{2}\ln \left( {\dfrac{{1 + x}}{{1 - x}}} \right)\]

Complete answer:First we have to evaluate the value of \[{\sinh ^{ - 1}}\left( {\dfrac{1}{2}} \right)\] by formula .
Putting the value of \[x\]equal to \[\dfrac{1}{2}\]we get
\[{\sinh ^{ - 1}}\dfrac{1}{2} = \ln \left( {\dfrac{1}{2} + \sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + 1} } \right)\]
 \[ = \ln \left( {\dfrac{1}{2} + \sqrt {\left( {\dfrac{1}{4}} \right) + 1} } \right)\]
\[ = \ln \left( {\dfrac{1}{2} + \sqrt {\dfrac{5}{4}} } \right)\]
So we get \[{\sinh ^{ - 1}}\left( {\dfrac{1}{2}} \right)\] \[ = \ln \left( {\dfrac{1}{2} + \sqrt {\dfrac{5}{4}} } \right)\]
Now we have to equate it with the value of \[{\tanh ^{ - 1}}x\]and after that with the help of a formula we will get the value of \[x\].
\[\dfrac{1}{2}\ln \left( {\dfrac{{1 + x}}{{1 - x}}} \right) = \ln \left( {\dfrac{1}{2} + \sqrt {\dfrac{5}{4}} } \right)\]
Now for the simplicity of calculation we will use some logarithmic formula and we can get
\[\dfrac{1}{2}\ln \left( {\dfrac{{1 + x}}{{1 - x}}} \right) = \ln {\left( {\dfrac{{1 + x}}{{1 - x}}} \right)^{\dfrac{1}{2}}}\]
Then putting these value in previous equation we get
\[\ln {\left( {\dfrac{{1 + x}}{{1 - x}}} \right)^{\dfrac{1}{2}}} = \ln \left( {\dfrac{1}{2} + \sqrt {\dfrac{5}{4}} } \right)\]
\[ \Rightarrow {\left( {\dfrac{{1 + x}}{{1 - x}}} \right)^{\dfrac{1}{2}}} = \left( {\dfrac{1}{2} + \sqrt {\dfrac{5}{4}} } \right)\]
Taking square on both side we get
\[ \Rightarrow \left( {\dfrac{{1 + x}}{{1 - x}}} \right) = {\left( {\dfrac{1}{2} + \sqrt {\dfrac{5}{4}} } \right)^2}\]
\[ \Rightarrow \left( {\dfrac{{1 + x}}{{1 - x}}} \right) = \left( {{{\left( {\dfrac{1}{2}} \right)}^2} + \dfrac{5}{4} + 2 \times \dfrac{1}{2} \times \dfrac{{\sqrt 5 }}{2}} \right)\]
After simplifying we get
\[ \Rightarrow \left( {\dfrac{{1 + x}}{{1 - x}}} \right) = \left( {\dfrac{3}{2} + \dfrac{{\sqrt 5 }}{2}} \right)\]
\[ \Rightarrow \left( {\dfrac{{1 + x}}{{1 - x}}} \right) = \left( {\dfrac{{3 + \sqrt 5 }}{2}} \right)\]
After cross multiplication we get
\[ \Rightarrow 2 \times \left( {1 + x} \right) = \left( {3 + \sqrt 5 } \right) \times \left( {1 - x} \right)\]
\[ \Rightarrow 2 + 2x = 3 - 3x + \sqrt 5 - \sqrt 5 x\]
After simplification and taking all \[x\] containing term in one side we get
\[ \Rightarrow 5x + \sqrt 5 x = 1 + \sqrt 5 \]
\[ \Rightarrow x \times \left( {5 + \sqrt 5 } \right) = \left( {1 + \sqrt 5 } \right)\]
After simplifying we get
\[ \Rightarrow x = \dfrac{{1 + \sqrt 5 }}{{5 + \sqrt 5 }}\]
To rationalise we multiply both numerator and denominator with the conjugate of denominator which is \[\left( {5 - \sqrt 5 } \right)\]
\[ \Rightarrow x = \dfrac{{\left( {1 + \sqrt 5 } \right)\left( {5 - \sqrt 5 } \right)}}{{\left( {5 + \sqrt 5 } \right)\left( {5 - \sqrt 5 } \right)}}\]
After multiplication we get
\[ \Rightarrow x = \dfrac{{5 - \sqrt 5 + 5\sqrt 5 - 5}}{{25 - 5}}\]
After simplification we get
\[ \Rightarrow x = \dfrac{{4\sqrt 5 }}{{20}}\]
\[ \Rightarrow x = \dfrac{1}{{\sqrt 5 }}\]
So we get the final answer \[{\sinh ^{ - 1}}\left( {\dfrac{1}{2}} \right)\]=\[{\tanh ^{ - 1}}\left( {\dfrac{1}{{\sqrt 5 }}} \right)\]
So option 2 is the correct answer .

Note:
Remember we are dealing with hyperbolic function not simple trigonometric function .
We have to keep in mind basic logarithmic and fraction formulas and students can simplify the equation in a way s/he can find easiest . For cross check the answer student can put the value \[x = \dfrac{1}{{\sqrt 5 }}\]in inverse hyperbolic formula given by \[{\tanh ^{ - 1}}x = \dfrac{1}{2}\ln \left( {\dfrac{{1 + x}}{{1 - x}}} \right)\].Alternative but tedious method is to check the value of all four option and see which one is equal to \[{\sinh ^{ - 1}}\left( {\dfrac{1}{2}} \right)\].