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Question: Find the domain of \(f\left( x \right)={{\cos }^{-1}}2x+{{\sin }^{-1}}x\) ....

Find the domain of f(x)=cos⁡−12x+sin⁡−1xf\left( x \right)={{\cos }^{-1}}2x+{{\sin }^{-1}}x .

Explanation

Solution

Hint: In this question we are given the function as an inverse trigonometric function. To solve the question first we need to find the domain of sin⁡−1x{{\sin }^{-1}}x and cos⁡−12x{{\cos }^{-1}}2x with the help of inverse trigonometric function in which sin⁡−1{{\sin }^{-1}} and cos⁡−1{{\cos }^{-1}} lies between -1 and 1.

Complete step-by-step answer:
We know that the value of sin⁡−1θ{{\sin }^{-1}}\theta always lies between -1 and 1 for any angle θ\theta . Therefore, sin⁡−1θ{{\sin }^{-1}}\theta will be defined on the domain -1 and 1.
i.e. −1≤θ≤1-1\le \theta \le 1 ……(1)
Now we will find the value of sin⁡−1x{{\sin }^{-1}}x by substituting θ\theta as xx in equation (1)
i.e, −1≤θ≤1-1\le \theta \le 1
⇒−1≤x≤1\Rightarrow -1\le x\le 1
Therefore, sin⁡−1x{{\sin }^{-1}}x lies between -1 and 1 . So, the domain of sin⁡−1x{{\sin }^{-1}}x is [−1,1]\left[ -1,1 \right] ……(2)
We also know that the value of cos⁡−1θ{{\cos }^{-1}}\theta always lies between -1 and 1
Same as equation (1)
Now we will find the value of cos⁡−12x{{\cos }^{-1}}2x by substituting θ\theta as 2x2x in equation (1)
i.e. −1≤θ≤1-1\le \theta \le 1
⇒−1≤2x≤1\Rightarrow -1\le 2x\le 1
By dividing ′2′'2' throughout the above inequality, we get –
⇒−12≤x≤12\Rightarrow -\dfrac{1}{2}\le x\le \dfrac{1}{2}
Therefore, cos⁡−12x{{\cos }^{-1}}2x lies between −12-\dfrac{1}{2} and 12\dfrac{1}{2} So, the domain of cos⁡−12x{{\cos }^{-1}}2x is [−12,12]\left[ -\dfrac{1}{2},\dfrac{1}{2} \right] …..(3)
Now, find the value of f(x)=cos⁡−12x+sin⁡−1xf\left( x \right)={{\cos }^{-1}}2x+{{\sin }^{-1}}x .By substituting equation (2) and (3) . in
cos⁡−12x+sin⁡−1x{{\cos }^{-1}}2x+{{\sin }^{-1}}x , we get –
cos⁡−12x+sin⁡−1x=[−12,12]∩[−1,1]{{\cos }^{-1}}2x+{{\sin }^{-1}}x=\left[ -\dfrac{1}{2},\dfrac{1}{2} \right]\cap \left[ -1,1 \right]
=[−12,12]=\left[ -\dfrac{1}{2},\dfrac{1}{2} \right].

Note-: Students should know what is the domain and range of trigonometric functions. The domain of a function is the specific set of values that the independent variable in a function can take on. The range is the resulting value that the dependent variable can have as x varies throughout the domain.
They should not get confused between domain and range of inverse trigonometric functions.