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Question: If we have function as \[f(x) = {\cos ^{ - 1}}x + {\cos ^{ - 1}}\left\\{ {\dfrac{x}{2} + \dfrac{1}{2...

If we have function as f(x) = {\cos ^{ - 1}}x + {\cos ^{ - 1}}\left\\{ {\dfrac{x}{2} + \dfrac{1}{2}\sqrt {3 - 3{x^2}} } \right\\} then
1). f(23)=π3f\left( {\dfrac{2}{3}} \right) = \dfrac{\pi }{3}
2). f(23)=2cos⁡−1(23)−π3f\left( {\dfrac{2}{3}} \right) = 2{\cos ^{ - 1}}\left( {\dfrac{2}{3}} \right) - \dfrac{\pi }{3}
3). f(13)=π3f\left( {\dfrac{1}{3}} \right) = \dfrac{\pi }{3}
4). f(13)=π3+2cos⁡−1(13)f\left( {\dfrac{1}{3}} \right) = \dfrac{\pi }{3} + 2{\cos ^{ - 1}}\left( {\dfrac{1}{3}} \right)

Explanation

Solution

We will have to substitute the value of x as x=cos⁡θx = \cos \theta in order to simplify the given expression. We will use various arithmetic properties and trigonometric identities like 1−cos⁡2θ=sin⁡2θ1 - {\cos ^2}\theta = {\sin ^2}\theta to simplify the expression and calculate the required value.

Complete step-by-step solution:
Trigonometric functions are also known as Circular Functions can be simply defined as the functions of an angle of a triangle. It means that the relationship between the angles and sides of a triangle are given by these trigonometric functions.
The angles of the sine, the cosine, and the tangent are the primary classification of functions of trigonometry. And the three functions which are the cotangent, the secant and the cosecant can be derived from the primary functions.
We are given the following expression f(x) = {\cos ^{ - 1}}x + {\cos ^{ - 1}}\left\\{ {\dfrac{x}{2} + \dfrac{1}{2}\sqrt {3 - 3{x^2}} } \right\\}
Let x=cos⁡θx = \cos \theta which gives us the value θ=cos⁡−1x\theta = {\cos ^{ - 1}}x
Therefore we get the following expression
f(x) = \theta + {\cos ^{ - 1}}\left\\{ {\dfrac{{\cos \theta }}{2} + \dfrac{1}{2}\sqrt {3 - 3{{\cos }^2}\theta } } \right\\}
On simplification we get the following expression
f(x) = \theta + {\cos ^{ - 1}}\left\\{ {\dfrac{{\cos \theta }}{2} + \dfrac{{\sqrt 3 }}{2}\sqrt {1 - {{\cos }^2}\theta } } \right\\}
We know that 1−cos⁡2θ=sin⁡2θ1 - {\cos ^2}\theta = {\sin ^2}\theta
Therefore we get the following expression
f(x) = \theta + {\cos ^{ - 1}}\left\\{ {\dfrac{{\cos \theta }}{2} + \dfrac{{\sqrt 3 }}{2}\sin \theta } \right\\}
Putting the values of 12=cos⁡π3\dfrac{1}{2} = \cos \dfrac{\pi }{3} and 32=sin⁡π3\dfrac{{\sqrt 3 }}{2} = \sin \dfrac{\pi }{3} we get the following expression
f(x) = \theta + {\cos ^{ - 1}}\left\\{ {\cos \dfrac{\pi }{3}\cos \theta + \sin \dfrac{\pi }{3}\sin \theta } \right\\}
Using the identity cos⁡Acos⁡B+sin⁡Asin⁡B=cos⁡(A−B)\cos A\cos B + \sin A\sin B = \cos \left( {A - B} \right)
f(x) = \theta + {\cos ^{ - 1}}\left\\{ {\cos \left( {\dfrac{\pi }{3} - \theta } \right)} \right\\}
On further simplification we get the following expression
f(x)=θ+π3−θ=π3f(x) = \theta + \dfrac{\pi }{3} - \theta = \dfrac{\pi }{3}
Therefore we get the following expression
f(13)=π3f\left( {\dfrac{1}{3}} \right) = \dfrac{\pi }{3} and f(23)=π3f\left( {\dfrac{2}{3}} \right) = \dfrac{\pi }{3}
Therefore option (1) and option (3) are the correct options.

Note: To solve such type of questions one must have a strong grip over the concepts of trigonometry, its related formulas and basic trignometric identities so as to simplify the expression obtained at each step of the calculation. We must do the calculations carefully and should recheck them in order to get the desired result correctly and avoid errors.