To determine whether the function \( f(x) = \sin x \) defined on the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\) is one-to-one (1-1) and onto (onto), we can follow these steps:
### Step 1: Analyze the function
The function given is \( f(x) = \sin x \). We need to analyze its behavior on the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
### Step 2: Determine if the function is one-to-one
To check if the function is one-to-one, we can find the derivative of the function:
\[
f'(x) = \cos x
\]
Next, we evaluate the derivative on the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\):
- At \( x = -\frac{\pi}{2} \), \( f'(-\frac{\pi}{2}) = \cos(-\frac{\pi}{2}) = 0 \)
- At \( x = 0 \), \( f'(0) = \cos(0) = 1 \)
- At \( x = \frac{\pi}{2} \), \( f'(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) = 0 \)
Since \( \cos x \) is positive in the interval \((- \frac{\pi}{2}, \frac{\pi}{2})\), we conclude that \( f'(x) > 0 \) for all \( x \) in \((- \frac{\pi}{2}, \frac{\pi}{2})\). Therefore, \( f(x) \) is strictly increasing on the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
### Step 3: Conclusion on one-to-one
Since \( f(x) \) is strictly increasing, it is one-to-one (1-1).
### Step 4: Determine if the function is onto
Next, we need to check if the function is onto. The range of \( f(x) = \sin x \) as \( x \) varies from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) is:
\[
f\left(-\frac{\pi}{2}\right) = \sin\left(-\frac{\pi}{2}\right) = -1
\]
\[
f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1
\]
Thus, as \( x \) varies from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), \( f(x) \) takes all values from \(-1\) to \(1\).
### Step 5: Conclusion on onto
Since the range of the function \( f(x) \) is \([-1, 1]\) and this matches the codomain \([-1, 1]\), we conclude that \( f(x) \) is onto.
### Final Conclusion
Since \( f(x) \) is both one-to-one and onto, we can conclude that:
\[
f: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \to [-1, 1] \text{ is a bijection (1-1 and onto).}
\]
