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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. he 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 .
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}
2θ=cos⁡−1(−32)2\theta = {\cos ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)
Now we need to find the value of cos⁡−1(−32){\cos ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right). From trigonometric table we know that,
cos⁡(π−π6)=−32\cos \left( {\pi - \dfrac{\pi }{6}} \right) = - \dfrac{{\sqrt 3 }}{2}
cos⁡(6π−π6)=cos⁡5π6=−32\cos \left( {\dfrac{{6\pi - \pi }}{6}} \right) = \cos \dfrac{{5\pi }}{6} = - \dfrac{{\sqrt 3 }}{2}
5π6=cos⁡−1(−32)\dfrac{{5\pi }}{6} = {\cos ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)
Thus substituting in above equation we get,
cos⁡2θ=cos⁡5π6\cos 2\theta=\cos \dfrac{{5\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±5π6  θ=nπ±5π12   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 5π12\dfrac{{5\pi }}{{12}},which lies between the range (−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right).
Thus the principal value is 5π12\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 (−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right) and the domain of inverse function of sine is [−1,1][ - 1,1].