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.The 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 \[\theta\].
Thus, \[{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right) = \theta \]
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.
$\cos 2\theta = 1 - \dfrac{{4 - \sqrt 3 }}{4} = \dfrac{{4 - 4 + 2\sqrt 3 }}{4}$
$\cos 2\theta = \dfrac{{2\sqrt 3 }}{4} = \dfrac{{\sqrt 3 }}{2}$
$\cos 2\theta = \dfrac{{\sqrt 3 }}{2}$
From trigonometric table we know that,\[\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}\]
Substituting in above equation we get,
$\cos 2\theta=\cos \dfrac{\pi }{6}$
The general solution of trigonometric function $\cos x = \cos \alpha$ is given as $\cos x=2n\pi \pm\cos\alpha$
Hence the equation becomes,
\[
  2\theta = 2n\pi \pm \dfrac{\pi }{6} \\
  \theta = n\pi \pm \dfrac{\pi }{{12}} \\
 \]
Thus we got the principal value of inverse sine function as \[\dfrac{\pi }{{12}}\],which lies between the range \[\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)\].
Thus the principal value is \[\dfrac{\pi }{{12}}\].

Note: We can also find the principal value of the above expression by using identities of sine function.
For the function, \[{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right)\], we can write \[\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\]as,
\[\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} = \left( {\dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }}} \right) - \left( {\dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }}} \right)\]
Now from the trigonometric table we know the value of these functions. Thus the above expression becomes,
$\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} = \left( {\dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }}} \right) - \left( {\dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }}} \right)$
$
= \left( {\sin \dfrac{\pi }{3} \times \cos \dfrac{\pi }{4}} \right) - \left( {\cos \dfrac{\pi }{3} \times \sin \dfrac{\pi }{4}} \right)$
Now this is of the form, \[\sin (A - B) = \sin A\cos B - \cos A\sin B\]
Thus we get,
$\left( {\sin \dfrac{\pi }{3} \times \cos \dfrac{\pi }{4}} \right) - \left( {\cos \dfrac{\pi }{3}\times \sin \dfrac{\pi }{4}} \right) = \sin \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right)$
$\sin \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right) = \sin \dfrac{{4\pi - 3\pi }}{{12}}$
$\sin \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{{12}}$
${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right) = {\sin ^{ - 1}}\sin \dfrac{\pi }{{12}}$
${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right) = \dfrac{\pi }{{12}}$