This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. The inverse sine function arcsin(x) is constructed by restricting sin(x) to the domain [-π/2, π/2] where sine is always increasing, making it one-to-one and therefore invertible; the resulting inverse function arcsin has domain [-1, 1] (the range of sin) and range [-π/2, π/2] (the restricted domain of sin). The domain of arcsin is [-1, 1] because these are the possible output values of the sine function (sin(x) always lies between -1 and 1). Choice B is correct because it correctly states the domain of the inverse as [-1, 1]. Choice A confuses the domain and range of arcsin, stating the domain is [-π/2, π/2] when actually that's the range—the domain of arcsin is [-1, 1] (the possible sine values). Key to inverse trig functions: remember that arcsin has domain [-1, 1] and range [-π/2, π/2], arccos has domain [-1, 1] and range [0, π], and arctan has domain all reals and range (-π/2, π/2)—the domain is always the possible trig output values, and the range is the restricted input interval.

This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. A function has an inverse only if it is one-to-one (each output corresponds to exactly one input, passing the horizontal line test), and since sine, cosine, and tangent are periodic functions that repeat values, we must restrict their domains to intervals where they are always increasing or always decreasing. Without restriction, sine takes the same value at multiple angles (for example, sin(π/6) = sin(5π/6) = 1/2), so given an output like 1/2, we couldn't determine a unique input—the function wouldn't pass the horizontal line test. By restricting to [-π/2, π/2], each output value occurs at exactly one input, making the function invertible. Choice A is correct because it properly identifies the one-to-one requirement. Choice B incorrectly claims that periodicity makes the function invertible, when in fact periodicity (repeating values) is exactly why we need to restrict the domain to create a one-to-one function. The restriction is necessary because a function can only have an inverse if it's one-to-one (passes the horizontal line test), and sine, cosine, and tangent are periodic (repeating), so they're only one-to-one on restricted intervals where they're always increasing or always decreasing.

This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. The inverse sine function arcsin(x) is constructed by restricting sin(x) to the domain [-π/2, π/2] where sine is always increasing, making it one-to-one and therefore invertible; the resulting inverse function arcsin has domain [-1, 1] (the range of sin) and range [-π/2, π/2] (the restricted domain of sin). The range of arcsin is [-π/2, π/2] because these are the angles we allow in the restricted domain of sine that we're inverting. Choice C is correct because it correctly gives the range of the inverse as [-π/2, π/2]. Choice A confuses the domain and range of arcsin, stating the range is [-1, 1] when actually that's the domain—the range of arcsin is [-π/2, π/2] (the restricted domain of sin). Key to inverse trig functions: remember that arcsin has domain [-1, 1] and range [-π/2, π/2], arccos has domain [-1, 1] and range [0, π], and arctan has domain all reals and range (-π/2, π/2)—the domain is always the possible trig output values, and the range is the restricted input interval.

This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. The inverse sine function arcsin(x) is constructed by restricting sin(x) to the domain [-π/2, π/2] where sine is always increasing, making it one-to-one and therefore invertible; the resulting inverse function arcsin has domain [-1, 1] (the range of sin) and range [-π/2, π/2] (the restricted domain of sin). For sin(arcsin(x)) where x ∈ [-1, 1], arcsin(x) gives an angle in [-π/2, π/2], and taking sine of that angle returns the original value x. However, arcsin(sin(x)) = x only when x is already in [-π/2, π/2]; for other values like x = 3π/4, arcsin(sin(3π/4)) gives π/4 (the angle in the restricted range with the same sine value). Choice D is correct because it correctly states the interval where the composition equals x as [-π/2, π/2]. Choice A confuses the domain of arcsin (which is [-1, 1] for the input to arcsin) with the values of x where arcsin(sin(x)) = x holds. For composition: f(f⁻¹(x)) = x always works (for x in the domain of f⁻¹), but f⁻¹(f(x)) = x only when x is in the restricted domain—outside the restricted interval, the inverse function returns an equivalent angle in the restricted range.

This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. The inverse cosine function arccos(x) uses the restriction of cos(x) to [0, π] where cosine is always decreasing (one-to-one), giving arccos a domain of [-1, 1] and range of [0, π]. The range of arccos is [0, π] because these are the angles we allow in the restricted domain of cosine that we're inverting. Choice C is correct because it correctly gives the range of the inverse as [0, π]. Choice B confuses the domain and range of arccos, stating the range is [-π/2, π/2] when actually that's the range for arcsin—the range of arccos is [0, π] (the restricted domain of cos). Key to inverse trig functions: remember that arcsin has domain [-1, 1] and range [-π/2, π/2], arccos has domain [-1, 1] and range [0, π], and arctan has domain all reals and range (-π/2, π/2)—the domain is always the possible trig output values, and the range is the restricted input interval.

This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. The inverse tangent function arctan(x) is constructed by restricting tan(x) to (-π/2, π/2) where tangent is always increasing, and since tangent's range on this interval is all real numbers, arctan has domain (-∞, ∞) and range (-π/2, π/2). The restriction to (-π/2, π/2) is chosen because on this interval, tangent is always increasing (never decreases): as x increases from -π/2 to π/2, tan(x) increases from -∞ to ∞ without ever going back down, ensuring one-to-one correspondence. Choice A is correct because it properly identifies the always increasing property that makes it one-to-one. Choice D claims that tangent is always decreasing on the restricted interval, but actually it's always increasing on (-π/2, π/2), which is what makes it one-to-one. The restriction is necessary because a function can only have an inverse if it's one-to-one (passes the horizontal line test), and sine, cosine, and tangent are periodic (repeating), so they're only one-to-one on restricted intervals where they're always increasing or always decreasing.

This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. The inverse cosine function $\arccos(x)$ uses the restriction of $\cos(x)$ to $[0, \pi]$ where cosine is always decreasing (one-to-one), giving $\arccos$ a domain of $[-1, 1]$ and range of $[0, \pi]$. Without restriction, cosine takes the same value at multiple angles (for example, $\cos(\pi/3) = \cos(-\pi/3) = 1/2$), so given an output like 1/2, we couldn't determine a unique input—the function wouldn't pass the horizontal line test. By restricting to $[0, \pi]$, each output value occurs at exactly one input, making the function invertible. Choice B is correct because it correctly states the restricted domain for $\arccos$ as $[0, \pi]$. Choice A confuses the restriction for arcsin (which is $[-\pi/2, \pi/2]$) with the restriction for $\arccos$ (which is $[0, \pi]$). The restriction is necessary because a function can only have an inverse if it's one-to-one (passes the horizontal line test), and sine, cosine, and tangent are periodic (repeating), so they're only one-to-one on restricted intervals where they're always increasing or always decreasing.

This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. The inverse sine function arcsin(x) is constructed by restricting sin(x) to the domain [-π/2, π/2] where sine is always increasing, making it one-to-one and therefore invertible; the resulting inverse function arcsin has domain [-1, 1] (the range of sin) and range [-π/2, π/2] (the restricted domain of sin). For composition: f(f⁻¹(x)) = x always works (for x in the domain of f⁻¹), but f⁻¹(f(x)) = x only when x is in the restricted domain—outside the restricted interval, the inverse function returns an equivalent angle in the restricted range. The restriction to [-π/2, π/2] is chosen because on this interval, sine is always increasing (never decreases): as x increases from -π/2 to π/2, sin(x) increases from -1 to 1 without ever going back down, ensuring one-to-one correspondence. Choice B is correct because it correctly states restricted domain. Choice A confuses the domain and range of arcsin, stating the domain is [-π/2, π/2] when actually that's the range—the domain of arcsin is [-1, 1] (the possible sine values). For composition: f(f⁻¹(x)) = x always works (for x in the domain of f⁻¹), but f⁻¹(f(x)) = x only when x is in the restricted domain—outside the restricted interval, the inverse function returns an equivalent angle in the restricted range.

This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. A function has an inverse only if it is one-to-one (each output corresponds to exactly one input, passing the horizontal line test), and since sine, cosine, and tangent are periodic functions that repeat values, we must restrict their domains to intervals where they are always increasing or always decreasing. For composition: f(f⁻¹(x)) = x always works (for x in the domain of f⁻¹), but f⁻¹(f(x)) = x only when x is in the restricted domain—outside the restricted interval, the inverse function returns an equivalent angle in the restricted range. To evaluate arcsin([value]), we ask: what angle in [-π/2, π/2] has sine equal to [value]? Using special triangle values, we know sin(π/[6 or 4 or 3]) = [value], and since π/[6 or 4 or 3] is in the range of arcsin, we have arcsin([value]) = π/[6 or 4 or 3]. Choice B is correct because it correctly evaluates using restricted range. Choice A gives a value outside the range of arcsin, stating arcsin(1/2) = [5π/6 or other], but arcsin must return an angle in [-π/2, π/2], and [wrong value] is outside this interval. For composition: f(f⁻¹(x)) = x always works (for x in the domain of f⁻¹), but f⁻¹(f(x)) = x only when x is in the restricted domain—outside the restricted interval, the inverse function returns an equivalent angle in the restricted range.

This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. The inverse sine function $\arcsin(x)$ is constructed by restricting $\sin(x)$ to the domain $[-\pi/2, \pi/2]$ where sine is always increasing, making it one-to-one and therefore invertible; the resulting inverse function arcsin has domain $[-1, 1]$ (the range of $\sin$) and range $[-\pi/2, \pi/2]$ (the restricted domain of $\sin$). The domain of arcsin is $[-1, 1]$ because these are the possible output values of the sine function ($\sin(x)$ always lies between -1 and 1). The range of arcsin is $[-\pi/2, \pi/2]$ because these are the angles we allow in the restricted domain of sine that we're inverting. Choice B is correct because it correctly states the domain of the inverse. Choice A confuses the domain and range of arcsin, stating the domain is $[-\pi/2, \pi/2]$ when actually that's the range—the domain of arcsin is $[-1, 1]$ (the possible sine values). Key to inverse trig functions: remember that arcsin has domain $[-1, 1]$ and range $[-\pi/2, \pi/2]$, arccos has domain $[-1, 1]$ and range $[0, \pi]$, and arctan has domain all reals and range $(-\pi/2, \pi/2)$—the domain is always the possible trig output values, and the range is the restricted input interval.