Question

q4 (1 point)
on which interval is $\sin(x)$ invertible? select all that apply.
$\left(-\frac{\pi}{2}, \frac{\pi}{2}\
ight)$
$\left-\frac{\pi}{2}, \frac{\pi}{2}\
ight$
$(0, \pi)$
$0, \pi$

Explanation:

Step1: Recall invertibility condition

A function is invertible on an interval if it is one-to-one (injective) there, meaning it passes the Horizontal Line Test (no horizontal line intersects the function more than once).

Step2: Analyze $\sin(x)$ on $(-\frac{\pi}{2}, \frac{\pi}{2})$

On this open interval, $\sin(x)$ is strictly increasing (derivative $\cos(x) > 0$ here), so it is one-to-one.

Step3: Analyze $\sin(x)$ on $[-\frac{\pi}{2}, \frac{\pi}{2}]$

On this closed interval, $\sin(x)$ is strictly increasing (derivative $\cos(x) \geq 0$, with $\cos(x)=0$ only at the endpoints), so it is one-to-one.

Step4: Analyze $\sin(x)$ on $(0, \pi)$

On this interval, $\sin(x)$ increases to $\frac{\pi}{2}$ then decreases. For example, $\sin(\frac{\pi}{6}) = \sin(\frac{5\pi}{6}) = \frac{1}{2}$, so it fails the Horizontal Line Test.

Step5: Analyze $\sin(x)$ on $[0, \pi]$

On this closed interval, $\sin(0) = \sin(\pi) = 0$, so it fails the Horizontal Line Test (not one-to-one).

Answer:

$(-\frac{\pi}{2}, \frac{\pi}{2})$, $[-\frac{\pi}{2}, \frac{\pi}{2}]$