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Question: Solve the given inverse trigonometric equation for x, \[\cos \left( 2{{\sin }^{-1}}x \right)=\dfrac{...

Solve the given inverse trigonometric equation for x, cos⁡(2sin⁡−1x)=19,x>0\cos \left( 2{{\sin }^{-1}}x \right)=\dfrac{1}{9},x>0.

Explanation

Solution

Hint: In this question, we first need to apply the inverse of cosine functions on both sides. Then by using the properties of inverse functions we need to convert the inverse of sine function into simplified form on the left hand side. Now, on the right hand side using the property we can convert the inverse of the cosine term on the right hand side to the inverse of the sine term. Then, on simplifying the equation obtained in x we get the result.
2sin⁡−1x=sin⁡−1(2x1−x2)2{{\sin }^{-1}}x={{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)
cos⁡−1x=sin⁡−11−x2{{\cos }^{-1}}x={{\sin }^{-1}}\sqrt{1-{{x}^{2}}}.

Complete step-by-step solution -
Now, from the given expression in the question we have
⇒cos⁡(2sin⁡−1x)=19\Rightarrow \cos \left( 2{{\sin }^{-1}}x \right)=\dfrac{1}{9}
Now, on applying inverse of cosine on both the sides we get,
⇒2sin⁡−1x=cos⁡−1(19)\Rightarrow 2{{\sin }^{-1}}x={{\cos }^{-1}}\left( \dfrac{1}{9} \right)
As we already know that the inverse of cosine function in terms of inverse of sine function is given by
cos⁡−1x=sin⁡−11−x2{{\cos }^{-1}}x={{\sin }^{-1}}\sqrt{1-{{x}^{2}}}
Now, the above equation can be further written as
⇒2sin⁡−1x=sin⁡−11−(19)2\Rightarrow 2{{\sin }^{-1}}x={{\sin }^{-1}}\sqrt{1-{{\left( \dfrac{1}{9} \right)}^{2}}}
Let us now further simplify and rewrite it as follows
⇒2sin⁡−1x=sin⁡−18081\Rightarrow 2{{\sin }^{-1}}x={{\sin }^{-1}}\sqrt{\dfrac{80}{81}}
Now, let us simplify the left hand side in the above expression using properties of inverse function as follow
2sin⁡−1x=sin⁡−1(2x1−x2)2{{\sin }^{-1}}x={{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)
Now, on using the above property we can rewrite the equation as
⇒sin⁡−1(2x1−x2)=sin⁡−18081\Rightarrow {{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)={{\sin }^{-1}}\sqrt{\dfrac{80}{81}}
Now, on equating the right hand and left hand sides we get,
⇒2x1−x2=8081\Rightarrow 2x\sqrt{1-{{x}^{2}}}=\sqrt{\dfrac{80}{81}}
Let us now do squaring on both sides to simplify it further
⇒4x2(1−x2)=8081\Rightarrow 4{{x}^{2}}\left( 1-{{x}^{2}} \right)=\dfrac{80}{81}
Now, on further simplification we get,
⇒x2(1−x2)=2081\Rightarrow {{x}^{2}}\left( 1-{{x}^{2}} \right)=\dfrac{20}{81}
Now, the right hand side can also be written as follows
⇒x2(1−x2)=4×581\Rightarrow {{x}^{2}}\left( 1-{{x}^{2}} \right)=\dfrac{4\times 5}{81}
Now, this can be further rewritten as
⇒x2(1−x2)=49×59\Rightarrow {{x}^{2}}\left( 1-{{x}^{2}} \right)=\dfrac{4}{9}\times \dfrac{5}{9}
Let us now write it further in the simplified form
⇒x2(1−x2)=49×(1−49)\Rightarrow {{x}^{2}}\left( 1-{{x}^{2}} \right)=\dfrac{4}{9}\times \left( 1-\dfrac{4}{9} \right)
Now, on comparing the right hand side and left hand side we get,
⇒x2=49\Rightarrow {{x}^{2}}=\dfrac{4}{9}
Let us now apply the square root on both the sides
⇒x2=49\Rightarrow \sqrt{{{x}^{2}}}=\sqrt{\dfrac{4}{9}}
∴x=23\therefore x=\dfrac{2}{3}
Hence, the value of x is 23\dfrac{2}{3}.

Note: Instead of applying the inverse of cosine function in the beginning we can also solve it by first simplifying the cosine term inside and then write it in terms of inverse of cosine and then use the property of inverse trigonometric function to solve it further. Both the methods give the same result.
It is important to note that while finding the value of x we should not neglect any of the terms while squaring and rearranging because it changes the corresponding equation and so the value of x accordingly.
Here we consider only one value of x though there are other possible values because it is already mentioned in the question that x is greater than 0.