Constructing Inverse Trigonometric Functions - Geometry
Card 1 of 30
Which inverse trig function returns an angle in $\left[0,\pi\right]$?
Which inverse trig function returns an angle in $\left[0,\pi\right]$?
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$\arccos(x)$. The arccosine function outputs angles in its principal range.
$\arccos(x)$. The arccosine function outputs angles in its principal range.
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What is the principal restricted domain used to define $y=\arccos(x)$?
What is the principal restricted domain used to define $y=\arccos(x)$?
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$\left[0,\pi\right]$. This interval makes $\cos(x)$ strictly decreasing and one-to-one.
$\left[0,\pi\right]$. This interval makes $\cos(x)$ strictly decreasing and one-to-one.
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Evaluate $\arctan\left(\tan\left(-\frac{\pi}{4}\right)\right)$.
Evaluate $\arctan\left(\tan\left(-\frac{\pi}{4}\right)\right)$.
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$-\frac{\pi}{4}$. $-\frac{\pi}{4}$ is in the principal range of arctangent.
$-\frac{\pi}{4}$. $-\frac{\pi}{4}$ is in the principal range of arctangent.
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What is the domain of $y=\arcsin(x)$?
What is the domain of $y=\arcsin(x)$?
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$\left[-1,1\right]$. The domain of an inverse function equals the range of the original function.
$\left[-1,1\right]$. The domain of an inverse function equals the range of the original function.
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Evaluate $\arctan\left(\tan\left(-\frac{\pi}{4}\right)\right)$.
Evaluate $\arctan\left(\tan\left(-\frac{\pi}{4}\right)\right)$.
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$ -\frac{\pi}{4} $. $ -\frac{\pi}{4} $ is in the principal range of arctangent.
$ -\frac{\pi}{4} $. $ -\frac{\pi}{4} $ is in the principal range of arctangent.
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On what interval is $\tan(x)$ strictly increasing and commonly restricted to define an inverse?
On what interval is $\tan(x)$ strictly increasing and commonly restricted to define an inverse?
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$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. On this interval, $\tan(x)$ increases from $-\infty$ to $\infty$ without repeating.
$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. On this interval, $\tan(x)$ increases from $-\infty$ to $\infty$ without repeating.
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On what interval is $\cos(x)$ strictly decreasing, making it invertible there?
On what interval is $\cos(x)$ strictly decreasing, making it invertible there?
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$\left[0,\pi\right]$. On this interval, $\cos(x)$ decreases from $1$ to $-1$ without repeating.
$\left[0,\pi\right]$. On this interval, $\cos(x)$ decreases from $1$ to $-1$ without repeating.
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Choose the correct statement: $\arccos(x)$ outputs angles in which interval?
Choose the correct statement: $\arccos(x)$ outputs angles in which interval?
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$\left[0,\pi\right]$. Arccosine outputs angles in its principal range from $0$ to $\pi$.
$\left[0,\pi\right]$. Arccosine outputs angles in its principal range from $0$ to $\pi$.
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What is the domain of $y=\arcsin(x)$?
What is the domain of $y=\arcsin(x)$?
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$\left[-1,1\right]$. The domain of an inverse function equals the range of the original function.
$\left[-1,1\right]$. The domain of an inverse function equals the range of the original function.
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What property must a function have on a restricted domain to have an inverse function?
What property must a function have on a restricted domain to have an inverse function?
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It must be one-to-one on that domain. One-to-one means each output corresponds to exactly one input.
It must be one-to-one on that domain. One-to-one means each output corresponds to exactly one input.
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What test guarantees a function is one-to-one on an interval and thus invertible there?
What test guarantees a function is one-to-one on an interval and thus invertible there?
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The horizontal line test on that interval. If no horizontal line intersects the graph twice, the function is one-to-one.
The horizontal line test on that interval. If no horizontal line intersects the graph twice, the function is one-to-one.
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What monotonic behavior on an interval ensures a function is one-to-one there?
What monotonic behavior on an interval ensures a function is one-to-one there?
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Always increasing or always decreasing. Monotonic functions are one-to-one since they never repeat outputs.
Always increasing or always decreasing. Monotonic functions are one-to-one since they never repeat outputs.
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What is the principal restricted domain used to define $y=\arcsin(x)$?
What is the principal restricted domain used to define $y=\arcsin(x)$?
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$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. This interval makes $\sin(x)$ strictly increasing and one-to-one.
$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. This interval makes $\sin(x)$ strictly increasing and one-to-one.
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What is the principal value of $\arctan\left(\tan\left(\frac{5\pi}{6}\right)\right)$?
What is the principal value of $\arctan\left(\tan\left(\frac{5\pi}{6}\right)\right)$?
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$-\frac{\pi}{6}$. $\tan(\frac{5\pi}{6}) = -\frac{1}{\sqrt{3}}$, so $\arctan(-\frac{1}{\sqrt{3}}) = -\frac{\pi}{6}$.
$-\frac{\pi}{6}$. $\tan(\frac{5\pi}{6}) = -\frac{1}{\sqrt{3}}$, so $\arctan(-\frac{1}{\sqrt{3}}) = -\frac{\pi}{6}$.
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What is the range of $y=\arcsin(x)$?
What is the range of $y=\arcsin(x)$?
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$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. The range of an inverse function equals the domain of the original function.
$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. The range of an inverse function equals the domain of the original function.
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What is the principal restricted domain used to define $y=\arccos(x)$?
What is the principal restricted domain used to define $y=\arccos(x)$?
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$\left[0,\pi\right]$. This interval makes $\cos(x)$ strictly decreasing and one-to-one.
$\left[0,\pi\right]$. This interval makes $\cos(x)$ strictly decreasing and one-to-one.
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What is the range of $y=\arccos(x)$?
What is the range of $y=\arccos(x)$?
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$\left[0,\pi\right]$. The range of an inverse function equals the domain of the original function.
$\left[0,\pi\right]$. The range of an inverse function equals the domain of the original function.
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What is the principal restricted domain used to define $y=\arctan(x)$?
What is the principal restricted domain used to define $y=\arctan(x)$?
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$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. This interval makes $\tan(x)$ strictly increasing and one-to-one.
$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. This interval makes $\tan(x)$ strictly increasing and one-to-one.
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What is the range of $y=\arctan(x)$?
What is the range of $y=\arctan(x)$?
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$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. The range of an inverse function equals the domain of the original function.
$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. The range of an inverse function equals the domain of the original function.
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Choose the correct statement: $\arcsin(x)$ is defined only for which $x$-values?
Choose the correct statement: $\arcsin(x)$ is defined only for which $x$-values?
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$x\in\left[-1,1\right]$. Arcsine is only defined for inputs between $-1$ and $1$.
$x\in\left[-1,1\right]$. Arcsine is only defined for inputs between $-1$ and $1$.
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What is the domain of $y=\arccos(x)$?
What is the domain of $y=\arccos(x)$?
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$\left[-1,1\right]$. The domain of an inverse function equals the range of the original function.
$\left[-1,1\right]$. The domain of an inverse function equals the range of the original function.
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What is the domain of $y=\arctan(x)$?
What is the domain of $y=\arctan(x)$?
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$\left(-\infty,\infty\right)$. The domain of an inverse function equals the range of the original function.
$\left(-\infty,\infty\right)$. The domain of an inverse function equals the range of the original function.
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What is the range of $\sin(x)$ on $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$?
What is the range of $\sin(x)$ on $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$?
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$\left[-1,1\right]$. The sine function achieves all values from $-1$ to $1$ on this restricted domain.
$\left[-1,1\right]$. The sine function achieves all values from $-1$ to $1$ on this restricted domain.
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What is the range of $\cos(x)$ on $\left[0,\pi\right]$?
What is the range of $\cos(x)$ on $\left[0,\pi\right]$?
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$\left[-1,1\right]$. The cosine function achieves all values from $-1$ to $1$ on this restricted domain.
$\left[-1,1\right]$. The cosine function achieves all values from $-1$ to $1$ on this restricted domain.
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Evaluate $\tan\left(\arctan(x)\right)$ for $x\in\left(-\infty,\infty\right)$.
Evaluate $\tan\left(\arctan(x)\right)$ for $x\in\left(-\infty,\infty\right)$.
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$x$. Composing a function with its inverse yields the original input.
$x$. Composing a function with its inverse yields the original input.
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Evaluate $\cos\left(\arccos(x)\right)$ for $x\in\left[-1,1\right]$.
Evaluate $\cos\left(\arccos(x)\right)$ for $x\in\left[-1,1\right]$.
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$x$. Composing a function with its inverse yields the original input.
$x$. Composing a function with its inverse yields the original input.
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Evaluate $\sin\left(\arcsin(x)\right)$ for $x\in\left[-1,1\right]$.
Evaluate $\sin\left(\arcsin(x)\right)$ for $x\in\left[-1,1\right]$.
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$x$. Composing a function with its inverse yields the original input.
$x$. Composing a function with its inverse yields the original input.
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What is the value of $\arctan(-\sqrt{3})$ in the principal range?
What is the value of $\arctan(-\sqrt{3})$ in the principal range?
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$-\frac{\pi}{3}$. $\tan(-\frac{\pi}{3}) = -\sqrt{3}$, so $\arctan(-\sqrt{3}) = -\frac{\pi}{3}$.
$-\frac{\pi}{3}$. $\tan(-\frac{\pi}{3}) = -\sqrt{3}$, so $\arctan(-\sqrt{3}) = -\frac{\pi}{3}$.
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What is the value of $\arctan(1)$ in the principal range?
What is the value of $\arctan(1)$ in the principal range?
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$\frac{\pi}{4}$. $\tan(\frac{\pi}{4}) = 1$, so $\arctan(1) = \frac{\pi}{4}$.
$\frac{\pi}{4}$. $\tan(\frac{\pi}{4}) = 1$, so $\arctan(1) = \frac{\pi}{4}$.
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What is the value of $\arccos\left(-\frac{\sqrt{2}}{2}\right)$ in the principal range?
What is the value of $\arccos\left(-\frac{\sqrt{2}}{2}\right)$ in the principal range?
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$\frac{3\pi}{4}$. $\cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2}$, so $\arccos(-\frac{\sqrt{2}}{2}) = \frac{3\pi}{4}$.
$\frac{3\pi}{4}$. $\cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2}$, so $\arccos(-\frac{\sqrt{2}}{2}) = \frac{3\pi}{4}$.
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