The domain of sin-1\(\left( {\frac{{x + 1}}{3}} \right)\)
Domain of Inverse Sine Function Explained
Understanding the domain of inverse trigonometric functions is a fundamental concept in mathematics. The question asks us to find the domain of $\sin^{-1}\left( {\frac{{x + 1}}{3}} \right)$.
Understanding Inverse Sine Domain
The inverse sine function, denoted as $\sin^{-1}(u)$ or arcsin($u$), is defined only for specific values of its argument $u$. For the output of $\sin^{-1}(u)$ to be a real number, the argument $u$ must lie within the closed interval from -1 to 1, inclusive. Mathematically, this is expressed as:
- $-1 \le u \le 1$
This is because the range of the sine function is $[-1, 1]$, and for the inverse function to exist, its input must come from the range of the original function.
Solving for the Domain of \(\sin^{-1}\left( {\frac{{x + 1}}{3}} \right)\)
In our given function, the argument $u$ is $\frac{{x + 1}}{3}$. Therefore, to find the domain of $\sin^{-1}\left( {\frac{{x + 1}}{3}} \right)$, we must ensure that this argument satisfies the condition for the inverse sine function:
\[ -1 \le \frac{{x + 1}}{3} \le 1 \]
Now, we need to solve this compound inequality for $x$.
Step-by-Step Inequality Solution:
- Multiply all parts of the inequality by 3:
To eliminate the denominator, we multiply all three parts of the inequality by 3. Since 3 is a positive number, the direction of the inequality signs does not change.
\[ -1 \times 3 \le \left( {\frac{{x + 1}}{3}} \right) \times 3 \le 1 \times 3 \]
This simplifies to:
\[ -3 \le x + 1 \le 3 \]
- Subtract 1 from all parts of the inequality:
To isolate $x$, we subtract 1 from all three parts of the inequality.
\[ -3 - 1 \le x + 1 - 1 \le 3 - 1 \]
This simplifies to:
\[ -4 \le x \le 2 \]
Conclusion on the Domain
The inequality $-4 \le x \le 2$ defines the set of all possible values for $x$ for which the function $\sin^{-1}\left( {\frac{{x + 1}}{3}} \right)$ is defined. In interval notation, this range is represented as a closed interval.
- The domain of $\sin^{-1}\left( {\frac{{x + 1}}{3}} \right)$ is \([-4, 2]\).
This means that $x$ can take any value from -4 to 2, including -4 and 2. This solution aligns with the concept of finding the valid input range for inverse trigonometric functions.
| Function | Domain |
|---|---|
| \(\sin^{-1}(u)\) | \([-1, 1]\) |
| \(\cos^{-1}(u)\) | \([-1, 1]\) |
| \(\tan^{-1}(u)\) | \((-\infty, \infty)\) or \(R\) |
By correctly applying the domain rule for $\sin^{-1}(u)$ and solving the resulting inequality, we find the required domain.
Important Questions from Inverse Trigonometric Functions
What is \(1+\sin ^2\left(\cos ^{-1}\left(\frac{3}{\sqrt{17}}\right)\right)\) equal to ?
What is 2 cot \(\left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right)\) equal to ?
Consider the following statements:
1. There exists \({\rm{\theta }} \in \left( { - \frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{2}} \right)\) for which tan -1 (tan θ) ≠ θ
2. \({\sin ^{ - 1}}\left( {\frac{1}{3}} \right) - {\sin ^{ - 1}}\left( {\frac{1}{5}} \right) = {\sin ^{ - 1}}\left( {\frac{{2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{{15}}} \right)\)
Which of the above statements is/are correct?
Consider the following statements:
1. \({\tan ^{ - 1}}{\rm{x}} + {\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right) = {\rm{\pi }}\)
2. There exist x, y ∈ [-1, 1], where x ≠ y such that sin -1 x + cos -1 \({\rm{y}} = \frac{{\rm{\pi }}}{2}\)
Which of the above statements is/are correct?The value of \({\rm{tan}}\left( {2{{\tan }^{ - 1}}\frac{1}{5} - \frac{\pi }{4}} \right)\) is
