Solveeit Logo
Login

Question

Question: Choose the correct option from the given options below by solving the following question: The valu...

Choose the correct option from the given options below by solving the following question:
The value of 2 cos⁡−1x2\,{\cos ^{ - 1}}x is:
A.  cos⁡−1(2x2+1)\,{\cos ^{ - 1}}\left( {2{x^2} + 1} \right)
B.  cos⁡−12x1+x2\,{\cos ^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}}
C.  cos⁡−1(2x2−1)\,{\cos ^{ - 1}}\left( {2{x^2} - 1} \right)
D.  tan⁡−12x1−x2\,{\tan ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}

Explanation

Solution

Take the general expression as a parameter. Take the formula which is suitable for the question and substitute the values in terms of the parameter in the obtained equations. Solve the equations until the final answer is revealed.

Complete step-by-step solution:
This problem can be solved by using parameters.
Given question:
2 cos⁡−1x2\,{\cos ^{ - 1}}x. Let us take the general term as the parameter, taking the constant away.
Let us consider,
 cos⁡−1x=t\,{\cos ^{ - 1}}x = t
Which implies, if we apply the trigonometric ratio cos⁡\cos on both the sides of the equation above, we get;
x=cos⁡tx = \cos t
Which can also be written as;
cos⁡t=x\cos t = x
Now, let us take the following steps to find the value of 2 cos⁡−1x2\,{\cos ^{ - 1}}x.
According to the trigonometric expansion, we can have;
⇒cos⁡2t=2cos⁡2t−1\Rightarrow \cos 2t = 2{\cos ^2}t - 1
Substituting cos⁡t=x\cos t = x in the above equation, we get;
cos⁡2t=2x2−1\cos 2t = 2{x^2} - 1
Now, let us apply the inverse of cos⁡\cos on both the sides of the equation. We get;
2t=cos⁡−1(2x2−1)2t = {\cos ^{ - 1}}\left( {2{x^2} - 1} \right)
We already have t=cos⁡−1xt = {\cos ^{ - 1}}x. Substituting this value in the above equation, we get;
2cos⁡−1x=cos⁡−1(2x2−1)2{\cos ^{ - 1}}x = {\cos ^{ - 1}}\left( {2{x^2} - 1} \right)

Therefore, the correct option is C.

Note: In trigonometric equations, the parameter can represent anything as a third party including a side or an angle. A parameter is nothing but a variable which is assigned to condense the complex term into a simple term to avoid any confusion of complexity in solving the problem. Usually, a parameter is considered as the variable tt in all the geometric and trigonometric cases. To eliminate a parameter, we have to find the relation between the parameter and the equation, and substitute the suitable values until eliminated.