Question

Find the principal value of \[{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right)\].

Answer 1

Hint: The range of \[{\sin ^{ - 1}}x\] is between \[\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)\]. Equate the expression to \[\theta \]. Rationalize the given value of multiplying the numerator and denominator by \[\sqrt 2 \]. Thus find the value of \[\sin \theta \]. Now apply the equation to the basic trigonometric identity of \[\cos 2\theta \]. Simplify the expression and get the principal value.

Complete step-by-step answer:
A principal value of a function is the value selected at a point in the domain of a multiple-valued function, chosen so that the function has a single value at the point. he principal value of \[{\sin ^{ - 1}}x\] for \[x > 0\] , is the length of the arc of a unit circle centred at the origin which subtends an angle at the centre whose sine is x. For this reason \[{\sin ^{ - 1}}x\] is also denoted by \[\operatorname{arcsinx} \].
The principal value of \[{\sin ^{ - 1}}x\] branches to,
\[{\sin ^{ - 1}}x \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)\].
Hence the principal value of the given function will be between the range\[\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)\] .
Now we have been given the function, \[{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right)\] ,for which we need to find the principal value.
Let us take the principal value of \[{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right)\]as \[\theta \].
\[{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right) = \theta \]
Now we can write it as,
\[\sin \theta = \left( {\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right)\]
Now let us rationalize the above expression by multiplying the numerator and denominator of the expression with \[\sqrt 2 \]. Thus we get,
$\sin \theta = \left( {\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right) \times \dfrac{{\sqrt 2 }}{{\sqrt 2}}$
$\sin \theta = \left( {\dfrac{{(\sqrt 3 + 1) \times \sqrt 2 }}{{2\sqrt 2 \times \sqrt 2 }}} \right) = \left( {\dfrac{{\sqrt 6 + \sqrt 2 }}{{2 \times 2}}} \right) = \left( {\dfrac{{\sqrt 6 + \sqrt 2 }}{4}} \right)$
$\sin \theta = \dfrac{{\sqrt 6 + \sqrt 2 }}{4}$
We know the basic trigonometric identity,
\[\cos 2\theta = 1 - 2{\sin ^2}\theta \]
Now let us substitute the value of \[{\sin ^2}\theta \] in the above expression.
\[{\sin ^2}\theta = \dfrac{{{{(\sqrt 6 + \sqrt 2 )}^2}}}{{{4^2}}}\]
We know the basic identity and use it to solve the following expression.
\[{(a + b)^2} = {a^2} + 2ab + {b^2}\]
${\sin ^2}\theta = \dfrac{{{{(\sqrt 6 + \sqrt 2 )}^2}}}{{{4^2}}} = \dfrac{{{{(\sqrt 6 )}^2} + 2 \times \sqrt 6 \times \sqrt 2 + {{(\sqrt 2 )}^2}}}{{16}}$
${\sin ^2}\theta = \dfrac{{6 + 2\sqrt {12} + 2}}{{16}} = \dfrac{{8 + 4\sqrt 3 }}{{16}} = 2\left( {\dfrac{{4 + 2\sqrt 3 }}{{16}}} \right) = \left( {\dfrac{{4 + 2\sqrt 3 }}{8}} \right)$
${\sin ^2}\theta = \left( {\dfrac{{4 + 2\sqrt 3 }}{8}} \right)$
Thus we got the required value of \[{\sin ^2}\theta \]. Now let us put this in the equation of \[\cos 2\theta \].
\[\cos 2\theta = 1 - 2 \times \left( {\dfrac{{4 + 2\sqrt 3 }}{8}} \right)\], now let us simplify it.
$\cos 2\theta = 1 - \dfrac{{4 + \sqrt 3 }}{4} = \dfrac{{4 - 4 - 2\sqrt 3 }}{4}$
$\cos 2\theta = \dfrac{{ - 2\sqrt 3 }}{4} = - \dfrac{{\sqrt 3 }}{2}$
$\cos 2\theta = - \dfrac{{\sqrt 3 }}{2}$
$2\theta = {\cos ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)$
Now we need to find the value of \[{\cos ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)\]. From trigonometric table we know that,
$\cos \left( {\pi - \dfrac{\pi }{6}} \right) = - \dfrac{{\sqrt 3 }}{2}$
$\cos \left( {\dfrac{{6\pi - \pi }}{6}} \right) = \cos \dfrac{{5\pi }}{6} = - \dfrac{{\sqrt 3 }}{2}$
$\dfrac{{5\pi }}{6} = {\cos ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)$
Thus substituting in above equation we get,
$\cos 2\theta=\cos \dfrac{{5\pi }}{6}$
The general solution of trigonometric function $\cos x = \cos \alpha$ is given as $\cos x=2n\pi \pm\cos\alpha$
Hence the equation becomes,
\[
  2\theta = 2n\pi \pm \dfrac{{5\pi }}{6} \ \\
  \theta = n\pi \pm \dfrac{{5\pi }}{{12}} \ \\
\]
Thus we got the principal value of inverse sine function as \[\dfrac{{5\pi }}{{12}}\],which lies between the range \[\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)\].
Thus the principal value is \[\dfrac{{5\pi }}{{12}}\].

Note: To solve a question like these you should be familiar with the domain and range of the sine functions as well as the domain and range of the inverse sine functions. For us the range of inverse sine function is \[\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)\] and the domain of inverse function of sine is \[[ - 1,1]\].