Discuss why this statement is incorrect: arcsin(sin(x)) = x for all x. (Fill in the blank using interval notation.) This statement is only correct if x is in the interval The range of arcsin(x) is restricted to this interval because sin(x) is over this interval.
The correct answer and explanation is :
The statement arcsin(sin(x)) = x for all x is incorrect because the inverse sine function, arcsin(x), only returns values within a specific range. The correct statement is:
[
\arcsin(\sin(x)) = x \quad \text{only if } x \text{ is in the interval } \mathbf{[-\frac{\pi}{2}, \frac{\pi}{2}]}.
]
Explanation:
The function sin(x) is periodic with a period of (2\pi), meaning it repeats its values infinitely. However, the arcsin(x) function, also called the inverse sine function, is defined to return values only in its principal range:
[
\text{Range of } \arcsin(x) = \left[-\frac{\pi}{2}, \frac{\pi}{2} \right].
]
This restriction exists because an inverse function must be one-to-one (bijective), meaning it has exactly one output for each input. Since the sine function is not one-to-one over all real numbers, we must restrict it to an interval where it is one-to-one. The interval ( \left[-\frac{\pi}{2}, \frac{\pi}{2} \right] ) is chosen because within this range, sin(x) is increasing and covers all possible sine values from (-1) to (1) exactly once.
If ( x ) is outside this interval, the equation (\arcsin(\sin(x)) = x) does not always hold. For example:
- If ( x = \frac{3\pi}{4} ), then sin(x) = \sin(\frac{\pi}{4}).
- However, arcsin(sin(x)) returns ( \frac{\pi}{4} ), not ( \frac{3\pi}{4} ), since (\frac{3\pi}{4}) is outside the principal range.
Thus, the function arcsin(sin(x)) only correctly returns ( x ) when ( x ) is already within ( [-\frac{\pi}{2}, \frac{\pi}{2}] ).