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

Find the principal value of sin⁡−1(3−122){\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right).

Explanation

Solution

Hint: The range of sin⁡−1x{\sin ^{ - 1}}x is between (−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right). Equate the expression to θ\theta . Rationalize the given value of multiplying the numerator and denominator by 2\sqrt 2 . Thus find the value of sin⁡θ\sin \theta . Now apply the equation to the basic trigonometric identity of cos⁡2θ\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⁡−1x{\sin ^{ - 1}}x for x>0x > 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⁡−1x{\sin ^{ - 1}}x is also denoted by arcsinx⁡\operatorname{arcsinx} .
The principal value of sin⁡−1x{\sin ^{ - 1}}x branches to,
sin⁡−1x∈(−π2,π2){\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(−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right) .
Now we have been given the function,
sin⁡−1(3−122){\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(3−122){\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right) as θ\theta.
Thus, sin⁡−1(3−122)=θ{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right) = \theta
Now we can write it as,
sin⁡θ=(3−122)\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 2\sqrt 2 . Thus we get,
sin⁡θ=(3−122)×22\sin \theta = \left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right) \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}
sin⁡θ=((3−1)×222×2)=(6−22×2)=(6−24)\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⁡θ=6−24\sin \theta = \dfrac{{\sqrt 6 - \sqrt 2 }}{4}
We know the basic trigonometric identity,
cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2{\sin ^2}\theta
Now let us substitute the value of sin⁡2θ{\sin ^2}\theta in the above expression.
sin⁡2θ=(6−2)242{\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=a2−2ab+b2{(a - b)^2} = {a^2} - 2ab + {b^2}
sin⁡2θ=(6−2)242=(6)2−2×6×2+(2)216{\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⁡2θ=6−212+216=8−4316=2(4−2316)=(4−238){\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⁡2θ=(4−238){\sin ^2}\theta = \left( {\dfrac{{4 - 2\sqrt 3 }}{8}} \right)
Thus we got the required value of sin⁡2θ{\sin ^2}\theta . Now let us put this in the equation of cos⁡2θ\cos 2\theta .
cos⁡2θ=1−2×(4−238)\cos 2\theta = 1 - 2 \times \left( {\dfrac{{4 - 2\sqrt 3 }}{8}} \right), now let us simplify it.
cos⁡2θ=1−4−34=4−4+234\cos 2\theta = 1 - \dfrac{{4 - \sqrt 3 }}{4} = \dfrac{{4 - 4 + 2\sqrt 3 }}{4}
cos⁡2θ=234=32\cos 2\theta = \dfrac{{2\sqrt 3 }}{4} = \dfrac{{\sqrt 3 }}{2}
cos⁡2θ=32\cos 2\theta = \dfrac{{\sqrt 3 }}{2}
From trigonometric table we know that,cos⁡π6=32\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}
Substituting in above equation we get,
cos⁡2θ=cos⁡π6\cos 2\theta=\cos \dfrac{\pi }{6}
The general solution of trigonometric function cos⁡x=cos⁡α\cos x = \cos \alpha is given as cos⁡x=2nπ±cos⁡α\cos x=2n\pi \pm\cos\alpha
Hence the equation becomes,

2θ=2nπ±π6 θ=nπ±π12  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 π12\dfrac{\pi }{{12}},which lies between the range (−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right).
Thus the principal value is π12\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(3−122){\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right), we can write 3−122\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}as,
3−122=(32×12)−(12×12)\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,
3−122=(32×12)−(12×12)\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)
=(sin⁡π3×cos⁡π4)−(cos⁡π3×sin⁡π4) = \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⁡Acos⁡B−cos⁡Asin⁡B\sin (A - B) = \sin A\cos B - \cos A\sin B
Thus we get,
(sin⁡π3×cos⁡π4)−(cos⁡π3×sin⁡π4)=sin⁡(π3−π4)\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⁡(π3−π4)=sin⁡4π−3π12\sin \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right) = \sin \dfrac{{4\pi - 3\pi }}{{12}}
sin⁡(π3−π4)=sin⁡π12\sin \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{{12}}
sin⁡−1(3−122)=sin⁡−1sin⁡π12{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right) = {\sin ^{ - 1}}\sin \dfrac{\pi }{{12}}
sin⁡−1(3−122)=π12{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right) = \dfrac{\pi }{{12}}