Constructing Inverse Trigonometric Functions - Precalculus

When constructing the inverse sine function $\arcsin(x)$ , we restrict $\sin(x)$ to the domain $[-\pi/2,\pi/2]$ . Why is this restriction necessary for the inverse function to exist?
Answer: Because restricting the domain makes $\sin(x)$ one-to-one (it passes the horizontal line test) on $[-\pi/2,\pi/2]$ . 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.
Given the restriction that defines $\arcsin(x)$ (inverse of $\sin(x)$ on $[-\pi/2,\pi/2]$ ), evaluate $\arcsin(\sqrt{3}/2)$ in radians.
Answer: $\pi/3$ 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). 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 A is correct because it correctly evaluates using restricted range. Choice B 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. To evaluate inverse trig functions at special values, use your knowledge of special angles: arcsin(1/2) = π/6, arcsin(√2/2) = π/4, arcsin(√3/2) = π/3, and remember the answer must lie in the appropriate restricted range.
Because $\arcsin(x)$ returns values only in the restricted range $[-\pi/2,\pi/2]$ , the composition $\arcsin(\sin(x))$ equals $x$ only on the restricted domain of sine. For what values of $x$ does $\arcsin(\sin(x))=x$ ?
Answer: $x\in[-\pi/2,\pi/2]$ 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 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 C is correct because it correctly identifies the interval where the composition equals x. Choice A confuses the domain of arcsin (which is [-1,1]) with the interval where arcsin(sin(x)) = x. 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. 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.
To define the inverse cosine function $\arccos(x)$ , we restrict $\cos(x)$ to the interval $[0,\pi]$ , where it is always decreasing and therefore one-to-one. What is the range of $\arccos(x)$ ?
Answer: $[0,\pi]$ 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 states the restricted domain for cosine, which becomes the range of arccos. Choice B confuses the restriction for arcsin (which is [-π/2, π/2]) with the restriction for arccos (which is [0, π]). 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. 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.
Consider the piecewise function $h(x) = \begin{cases} \sin(x) & \text{if } x \in [0, \frac{\pi}{2}] \\ \cos(x) & \text{if } x \in (\frac{\pi}{2}, \pi] \end{cases}$ . Which statement correctly describes why this function does or does not have an inverse?
Answer: No inverse exists because both pieces achieve the same output values The function has no inverse because it fails the horizontal line test. Since both sin(x) on [0, π/2] and cos(x) on (π/2, π] can achieve the same output values (for example, sin(π/6) = 1/2 and cos(π/3) = 1/2), the function is not one-to-one across its entire domain. Choice A is incorrect because continuity isn't required for inverse existence, only the one-to-one property. Choice C is incorrect because individual monotonicity of pieces doesn't guarantee global one-to-one property. Choice D is incorrect because being defined on a closed interval doesn't ensure invertibility.
For the inverse function, evaluate $\sin(\arcsin(1/2))$ .
Answer: $1/2$ 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.
A function $g(x) = 2\sin(3x) + 1$ is defined on the interval $\left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$ . To verify that an inverse function can be constructed, which property must be confirmed about $g(x)$ on this interval?
Answer: The function must be either strictly increasing or strictly decreasing throughout For an inverse to exist, g(x) must be one-to-one on the given interval, which requires it to be either strictly increasing or strictly decreasing throughout the interval. Since g'(x) = 6cos(3x) and cos(3x) > 0 for x ∈ [-π/6, π/6] (as 3x ∈ [-π/2, π/2]), the function is strictly increasing. Choice A is incorrect because extrema location doesn't determine invertibility. Choice C is incorrect because having exactly one horizontal tangent would actually indicate the function changes from increasing to decreasing. Choice D is incorrect because symmetry would violate the one-to-one property.
Compare the standard domain restrictions used to define $\arcsin(x)$ and $\arccos(x)$ . Why are different restrictions used for $\sin(x)$ versus $\cos(x)$ when defining their inverse functions?
Answer: Because the restrictions are chosen so each function is one-to-one on a continuous interval; $\sin(x)$ is monotonic on $[-\pi/2,\pi/2]$ and $\cos(x)$ is monotonic on $[0,\pi]$ . 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. 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. 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 B is correct because it properly identifies one-to-one requirement. Choice D 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.
Given the restricted definition of inverse sine where $\arcsin(x)$ has range $[-\pi/2,\pi/2]$ , what is $\arcsin\left(\tfrac{1}{2}\right)$ (in radians)?
Answer: $\pi/6$ 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). To evaluate arcsin(1/2), we ask: what angle in [-π/2, π/2] has sine equal to 1/2? Using special triangle values, we know sin(π/6) = 1/2, and since π/6 is in the range of arcsin, we have arcsin(1/2) = π/6. Choice A is correct because it correctly evaluates using the restricted range. Choice B gives a value outside the range of arcsin, stating arcsin(1/2) = 5π/6, but arcsin must return an angle in [-π/2, π/2], and 5π/6 is outside this interval. To evaluate inverse trig functions at special values, use your knowledge of special angles: arcsin(1/2) = π/6, arcsin(√2/2) = π/4, arcsin(√3/2) = π/3, and remember the answer must lie in the appropriate restricted range.
Given that $\sin(x)$ is restricted to $[-\pi/2,\pi/2]$ to define $\arcsin(x)$ , what is the range of $\arcsin(x)$ ?
Answer: $[-\pi/2,\pi/2]$ 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.
For what values of $x$ does $\arcsin(\sin(x))=x$ hold, given that $\arcsin(x)$ returns values only in its standard range?
Answer: $x\in[-\pi/2,\pi/2]$ 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.
Given that $\cos(x)$ is restricted to $[0,\pi]$ so it is one-to-one, the inverse function is $\arccos(x)$ . What is the range of $\arccos(x)$ ?
Answer: $[0,\pi]$ 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 $[0, \pi]$ . 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 C is correct because it correctly gives the range of the inverse. Choice B confuses the restriction for arcsin (which is $[-\pi/2, \pi/2]$ ) with the restriction for arccos (which is $[0, \pi]$ ). 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.
When constructing $\arccos(x)$ , we restrict $\cos(x)$ to $[0,\pi]$ so it is one-to-one. Which interval enables construction of the inverse cosine function $\arccos(x)$ ?
Answer: $[0,\pi]$ 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]. 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 C is correct because it correctly states restricted domain. Choice A confuses the restriction for arcsin (which is [-π/2, π/2]) with the restriction for arccos (which is [0, π]). 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.
A student constructs an inverse for $y = \sin(x)$ by restricting the domain to $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$ . On this interval, $\sin(x)$ decreases from $1$ to $-1$ then increases back to $0$ . What fundamental requirement for inverse functions is violated?
Answer: The function must be strictly monotonic, but $\sin(x)$ both decreases and increases When determining if a function has an inverse, you need to check if it's one-to-one (bijective). This means each output value corresponds to exactly one input value, which requires the function to be strictly monotonic—either always increasing or always decreasing throughout its domain. Looking at $\sin(x)$ on $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$ , the function starts at 1, decreases to -1 at $x = \pi$ , then increases back to 0. This creates a problem: the function takes on the same y-values (like $y = 0$ ) at multiple x-values within this interval. For instance, $\sin\left(\frac{\pi}{2} + \epsilon\right) = \sin\left(\frac{3\pi}{2} - \epsilon\right)$ for small positive $\epsilon$ . This violates the one-to-one requirement because an inverse function wouldn't know which x-value to return for these repeated y-values. Answer C correctly identifies that the function must be strictly monotonic, which $\sin(x)$ fails to be on this interval since it both decreases and increases. Answer A is wrong because $\sin(x)$ is continuous everywhere, including at $x = \pi$ . Answer B is incorrect since $\sin(x)$ has a finite range $[-1,1]$ and never approaches infinity. Answer D is false because $\sin(x)$ is differentiable everywhere and has no corners—its derivative exists at every point. Remember: for inverse functions, look for strict monotonicity first. The standard domain restriction for $\arcsin$ uses $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ precisely because $\sin(x)$ is strictly increasing there.
Consider the function $f(x) = \cos(x)$ on the interval $[\pi, 2\pi]$ . A student claims this restriction allows for the construction of an inverse function. Which statement best evaluates this claim?
Answer: Correct, because $\cos(x)$ is strictly decreasing on $[\pi, 2\pi]$ The claim is correct. On the interval [π, 2π], cos(x) is strictly decreasing, going from cos(π) = -1 to cos(2π) = 1. Since the function is strictly monotonic (decreasing) on this interval, it is one-to-one and therefore has an inverse. Choice A incorrectly states that cos(x) is increasing on this interval. Choice C misunderstands that the range being [-1, 1] doesn't prevent inverse function construction. Choice D incorrectly suggests additional range restriction is needed when the monotonic property already ensures invertibility.
To construct $\arccos(x)$ , we restrict $\cos(x)$ to $[0,\pi]$ , where it is always decreasing and one-to-one. Which interval enables construction of the inverse cosine function?
Answer: $[0,\pi]$ 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 restriction to [0, π] is chosen because on this interval, cosine is always decreasing (never increases): as x increases from 0 to π, cos(x) decreases from 1 to -1 without ever going back up, ensuring one-to-one correspondence. Choice B is correct because it correctly states the restricted domain for cosine. Choice A confuses the restriction for arcsin (which is [-π/2, π/2]) with the restriction for arccos (which is [0, π]). 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.
A student creates a modified arcsine function by restricting $\sin(x)$ to the domain $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$ and argues it's valid because the function is 'mostly decreasing.' Which mathematical principle does this violate?
Answer: Inverse functions require strict monotonicity, not approximate or partial monotonicity When creating inverse functions, you're essentially "undoing" a function by swapping its inputs and outputs. For this to work mathematically, the original function must pass a crucial test: it must be strictly monotonic (either always increasing or always decreasing) on its restricted domain. The correct answer is D because "mostly decreasing" isn't sufficient. On the interval $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$ , $\sin(x)$ decreases from $\frac{\pi}{2}$ to $\pi$ , but then increases from $\pi$ to $\frac{3\pi}{2}$ . This creates a problem: multiple x-values produce the same y-value. For instance, both $x = \frac{2\pi}{3}$ and $x = \frac{4\pi}{3}$ give $\sin(x) = \frac{\sqrt{3}}{2}$ . An inverse function couldn't determine which x-value to return, violating the definition of a function. Answer A is incorrect because boundedness isn't the issue—many valid inverse functions have unbounded domains. Answer B is wrong because inverse functions don't require positive ranges; the standard arcsine has range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , which includes negative values. Answer C is false because symmetry about the origin isn't required—the standard arcsine domain $[-1,1]$ is symmetric, but this isn't a general requirement. Remember: for inverse functions, strict monotonicity is non-negotiable. "Mostly" or "approximately" monotonic won't work—even one local maximum or minimum in your restricted domain will create the multiple-input problem that breaks the inverse relationship.
When constructing the inverse tangent function $\arctan(x)$ , we restrict $\tan(x)$ to the interval $(-\pi/2,\pi/2)$ so that it is one-to-one (always increasing). Which interval enables construction of the inverse tangent function?
Answer: $(-\pi/2,\pi/2)$ 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 B is correct because it correctly states the restricted domain for tangent. Choice C uses the wrong restricted interval, suggesting [-π/2, π/2] when the standard restriction for tangent is the open interval (-π/2, π/2) to avoid the asymptotes at the endpoints. 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.
When constructing the inverse sine function $\arcsin(x)$ , we restrict $\sin(x)$ to the domain $[-\pi/2,\pi/2]$ , where it is always increasing. Why is this restriction necessary for the inverse function to exist?
Answer: Because $\sin(x)$ must be one-to-one (pass the horizontal line test) on the restricted domain to have an inverse function. 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 D 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. 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.
When constructing the inverse tangent function $\arctan(x)$ , we restrict $\tan(x)$ to the interval $(-\pi/2,\pi/2)$ where it is always increasing. What is the domain of $\arctan(x)$ ?
Answer: $(-\infty,\infty)$ 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 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 [-π/2, π/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 gives domain/range of inverse. Choice C confuses the restriction for arcsin (which is [-π/2, π/2]) with the restriction for arccos (which is [0, π]). 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.

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