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Question: Solve the following inverse trigonometric function: \({{\sin }^{-1}}x+{{\sin }^{-1}}\sqrt{1-{{x}^{...

Solve the following inverse trigonometric function:
sin⁡−1x+sin⁡−11−x2{{\sin }^{-1}}x+{{\sin }^{-1}}\sqrt{1-{{x}^{2}}}

Explanation

Solution

We have been asked to solve the trigonometric expression which is written as sin⁡−1x+sin⁡−11−x2{{\sin }^{-1}}x+{{\sin }^{-1}}\sqrt{1-{{x}^{2}}}. As you can carefully look at this expression then you will find that this expression is written in the form of sin⁡−1x+sin⁡−1y{{\sin }^{-1}}x+{{\sin }^{-1}}y which is the inverse trigonometric identity and is equal to sin⁡−1(x1−y2+y1−x2){{\sin }^{-1}}\left( x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}} \right). Then substitute x in place of x and 1−x2\sqrt{1-{{x}^{2}}} in place of y in this formula and then solve.

Complete step-by-step answer:
We have been given in the above problem the following inverse trigonometric expression:
sin⁡−1x+sin⁡−11−x2{{\sin }^{-1}}x+{{\sin }^{-1}}\sqrt{1-{{x}^{2}}}
And we have to solve the above expression. If you revise the concepts of inverse trigonometric identities then you will find that the above expression is written in the form of sin⁡−1x+sin⁡−1y{{\sin }^{-1}}x+{{\sin }^{-1}}y which is equal to:
sin⁡−1(x1−y2+y1−x2){{\sin }^{-1}}\left( x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}} \right)
On comparing the given expression to this formula we will find that x is equal to x and y in the formula is equal to 1−x2\sqrt{1-{{x}^{2}}} so substituting the value of y as 1−x2\sqrt{1-{{x}^{2}}} in the above expression we get,
sin⁡−1(x1−(1−x2)2+1−x21−x2){{\sin }^{-1}}\left( x\sqrt{1-{{\left( \sqrt{1-{{x}^{2}}} \right)}^{2}}}+\sqrt{1-{{x}^{2}}}\sqrt{1-{{x}^{2}}} \right)
In the above expression, 1−x2\sqrt{1-{{x}^{2}}} has written twice so we can put 1−x2\sqrt{1-{{x}^{2}}} to the power 2.
sin⁡−1(x1−(1−x2)2+(1−x2)2){{\sin }^{-1}}\left( x\sqrt{1-{{\left( \sqrt{1-{{x}^{2}}} \right)}^{2}}}+{{\left( \sqrt{1-{{x}^{2}}} \right)}^{2}} \right)
In the above expression, we can write:
1−x2=(1−x2)12\sqrt{1-{{x}^{2}}}={{\left( 1-{{x}^{2}} \right)}^{\dfrac{1}{2}}}
sin⁡−1(x1−(1−x2)12(2)+(1−x2)12(2)){{\sin }^{-1}}\left( x\sqrt{1-{{\left( 1-{{x}^{2}} \right)}^{\dfrac{1}{2}}}^{\left( 2 \right)}}+{{\left( 1-{{x}^{2}} \right)}^{\dfrac{1}{2}}}^{\left( 2 \right)} \right)
Now, 2 will be cancelled out in the power of (1−x2)\left( 1-{{x}^{2}} \right) and we get,
sin⁡−1(x1−(1−x2)+(1−x2)) =sin⁡−1(x1−1+x2+(1−x2)) \begin{aligned} & {{\sin }^{-1}}\left( x\sqrt{1-\left( 1-{{x}^{2}} \right)}+\left( 1-{{x}^{2}} \right) \right) \\\ & ={{\sin }^{-1}}\left( x\sqrt{1-1+{{x}^{2}}}+\left( 1-{{x}^{2}} \right) \right) \\\ \end{aligned}
In the above expression, +1 and -1 will be cancelled out and we are left with:
sin⁡−1(x0+x2+(1−x2)) =sin⁡−1(x(x2)12+(1−x2)) =sin⁡−1(x(x)+(1−x2)) \begin{aligned} & {{\sin }^{-1}}\left( x\sqrt{0+{{x}^{2}}}+\left( 1-{{x}^{2}} \right) \right) \\\ & ={{\sin }^{-1}}\left( x{{\left( {{x}^{2}} \right)}^{\dfrac{1}{2}}}+\left( 1-{{x}^{2}} \right) \right) \\\ & ={{\sin }^{-1}}\left( x\left( x \right)+\left( 1-{{x}^{2}} \right) \right) \\\ \end{aligned}
Now, multiplication of x with itself will give x2{{x}^{2}} in the above expression.
sin⁡−1(x2+(1−x2)) =sin⁡−1(x2+1−x2) =sin⁡−1(1) \begin{aligned} & {{\sin }^{-1}}\left( {{x}^{2}}+\left( 1-{{x}^{2}} \right) \right) \\\ & ={{\sin }^{-1}}\left( {{x}^{2}}+1-{{x}^{2}} \right) \\\ & ={{\sin }^{-1}}\left( 1 \right) \\\ \end{aligned}
From the inverse trigonometric ratios, we know that the value of sin⁡−1(1)=π2{{\sin }^{-1}}\left( 1 \right)=\dfrac{\pi }{2}.
π2\dfrac{\pi }{2}
Hence, we have solved the expression given above and the value we are getting is π2\dfrac{\pi }{2}.

Note: You can check the solution that you are getting is correct or not by putting the values of x in the given expression and then check whether solving them will given us the value of π2\dfrac{\pi }{2} or not.
The given expression is:
sin⁡−1x+sin⁡−11−x2{{\sin }^{-1}}x+{{\sin }^{-1}}\sqrt{1-{{x}^{2}}}
Substituting x as 1 in the above equation we get,
sin⁡−11+sin⁡−11−12 =sin⁡−11+sin⁡−10 =π2+0 =π2 \begin{aligned} & {{\sin }^{-1}}1+{{\sin }^{-1}}\sqrt{1-{{1}^{2}}} \\\ & ={{\sin }^{-1}}1+{{\sin }^{-1}}0 \\\ & =\dfrac{\pi }{2}+0 \\\ & =\dfrac{\pi }{2} \\\ \end{aligned}
The expression given above on substituting the value of x equals 1 gives the answer π2\dfrac{\pi }{2} which is the same as that we have solved above.
Now, if we put x as 12\dfrac{1}{\sqrt{2}} in the given expression we get,
sin⁡−112+sin⁡−11−(12)2 =sin⁡−112+sin⁡−11−(12) =sin⁡−112+sin⁡−1(12) =2sin⁡−112 \begin{aligned} & {{\sin }^{-1}}\dfrac{1}{\sqrt{2}}+{{\sin }^{-1}}\sqrt{1-{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}} \\\ & ={{\sin }^{-1}}\dfrac{1}{\sqrt{2}}+{{\sin }^{-1}}\sqrt{1-\left( \dfrac{1}{2} \right)} \\\ & ={{\sin }^{-1}}\dfrac{1}{\sqrt{2}}+{{\sin }^{-1}}\sqrt{\left( \dfrac{1}{2} \right)} \\\ & =2{{\sin }^{-1}}\dfrac{1}{\sqrt{2}} \\\ \end{aligned}
And we know that, sin⁡−112=π4{{\sin }^{-1}}\dfrac{1}{\sqrt{2}}=\dfrac{\pi }{4} so using this relation in the above equation we get,
2π4 =π2 \begin{aligned} & 2\dfrac{\pi }{4} \\\ & =\dfrac{\pi }{2} \\\ \end{aligned}
Again, we are getting the same answer that we have solved above.
Hence, we have checked that the solution of the given expression that we are getting is correct.