True or False? Between consecutive asymptotes, cotangent behaves in the same way as tangent.

False. Cotangent is decreasing across the whole interval between consecutive asymptotes, where tangent is increasing. The two do share a range of all real numbers, so the difference between them is entirely in the direction of travel.

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Inverse & Reciprocal Trigonometric Functions (College Board AP® Precalculus): Flashcards

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FrontInverse Trigonometric Functions
What do an inverse trigonometric function's inputs and outputs represent?

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What do an inverse trigonometric function's inputs and outputs represent?
Its input is a numerical value and its output is an angle, which is the reverse of a trigonometric function, whose input is an angle and whose output is a value.
So since $\sin \frac{π}{6} = \frac{1}{2}$ , arcsine takes an input of $\frac{1}{2}$ and returns $\frac{π}{6}$ .
True or False?
$\sin^{- 1} x$ is another way of writing $\frac{1}{\sin x}$ .
False.
$\sin^{- 1} x$ means the angle whose sine is $x$ , so the $- 1$ names the inverse function rather than acting as an exponent.
The reciprocal $\frac{1}{\sin x}$ has its own separate name in trigonometry, $\csc x$ , which is exactly why the two must not be confused.
Why does a trigonometric function need a restricted domain before it can have an inverse?
Because it is periodic, so every output value is produced by infinitely many different inputs, and $\sin \frac{π}{6}$ and $\sin \frac{5 π}{6}$ are both $\frac{1}{2}$ .
Restricting the domain to an interval on which the function takes each of its values exactly once makes it invertible.
Fill in the two gaps in the ranges of the inverse trigonometric functions.
Arcsine returns an angle in $[- \frac{π}{2}, \frac{π}{2}]$ and arctangent one in the open interval $(- \frac{π}{2}, \frac{π}{2})$ , while arccosine returns one in $[______,______]$ instead.
The completed sentence is:
Arcsine returns an angle in $[- \frac{π}{2}, \frac{π}{2}]$ and arctangent one in the open interval $(- \frac{π}{2}, \frac{π}{2})$ , while arccosine returns one in $[0, π]$ instead.
Each range is the restricted domain of the original function, and cosine needs a different one because it is decreasing on $[0, π]$ while taking every value from $- 1$ to $1$ exactly once.
What are the domains of the three inverse trigonometric functions?
Arcsine and arccosine both have domain $[- 1, 1]$ , because those are the only values sine and cosine ever produce.
Arctangent has domain all real numbers, because tangent takes every real value somewhere.
How is the exact value of $\sin^{- 1} (- \frac{\sqrt{3}}{2})$ found without a calculator?
Ask which angle in $[- \frac{π}{2}, \frac{π}{2}]$ has a sine of $- \frac{\sqrt{3}}{2}$ .
From the unit circle $\sin \frac{π}{3} = \frac{\sqrt{3}}{2}$ , and sine is odd, so the answer is $- \frac{π}{3}$ , which does lie in that interval.
True or False?
$\sin^{- 1} (\frac{1}{2}) = \frac{5 π}{6}$ , because $\sin \frac{5 π}{6} = \frac{1}{2}$ .
False.
The reason given is perfectly true, but $\frac{5 π}{6}$ is not in the range of arcsine, so it cannot be an output of that function.
The answer is $\frac{π}{6}$ , the one angle in the range whose sine is $\frac{1}{2}$ .
How is a composition such as $\sin (\cos^{- 1} \frac{1}{2})$ evaluated?
Work from the inside out.
First $\cos^{- 1} \frac{1}{2} = \frac{π}{3}$ , the angle in arccosine's range whose cosine is $\frac{1}{2}$ , and then $\sin \frac{π}{3} = \frac{\sqrt{3}}{2}$ .
True or False?
The graph of $y = \arctan x$ has horizontal asymptotes.
True.
They sit at $y = \frac{π}{2}$ and at $y = - \frac{π}{2}$ , the two ends of arctangent's range, which the curve approaches without ever reaching.
They are what the vertical asymptotes of tangent turn into once the graph is reflected to give the inverse.
Fill in the three gaps in the definitions of the reciprocal trigonometric functions.
$\sec θ = \frac{1}{______}, \csc θ = \frac{1}{______}, \cot θ = \frac{1}{______}$
The completed definitions are:
$\sec θ = \frac{1}{\cos θ}, \csc θ = \frac{1}{\sin θ}, \cot θ = \frac{1}{\tan θ}$
Each one is the reciprocal of its partner function, and each is undefined wherever that partner takes the value zero.
Cotangent can equivalently be written as $\frac{\cos θ}{\sin θ}$ , which is often the more convenient form.
Where does the graph of $y = \sec θ$ have vertical asymptotes?
Wherever $\cos θ = 0$ , so at $θ = \frac{π}{2} + k π$ for every integer $k$ .
A reciprocal is undefined exactly where the function it divides by takes the value zero.
Where do the graphs of $y = \csc θ$ and $y = \cot θ$ have vertical asymptotes?
Both have them at $θ = k π$ for every integer $k$ , which are the values where $\sin θ = 0$ .
For cosecant that follows straight from its definition, and for cotangent it follows from writing it as a quotient with sine underneath.
True or False?
Between consecutive asymptotes, cotangent behaves in the same way as tangent.
False.
Cotangent is decreasing across the whole interval between consecutive asymptotes, where tangent is increasing.
The two do share a range of all real numbers, so the difference between them is entirely in the direction of travel.
Why can secant and cosecant never take a value strictly between minus one and one?
Because they are reciprocals of sine and cosine, whose outputs never exceed $1$ in absolute value.
Taking the reciprocal of a number whose absolute value is at most $1$ gives one whose absolute value is at least $1$ , so the range of both is $(- \infty, - 1] \cup [1, \infty)$ .
How is the exact value of $\sec \frac{2 π}{3}$ found?
Find the value of the partner function first, then take its reciprocal.
Here $\cos \frac{2 π}{3} = - \frac{1}{2}$ , so $\sec \frac{2 π}{3} = \frac{1}{- \frac{1}{2}} = - 2$ .
What are the periods of secant, cosecant and cotangent?
Secant and cosecant both have period $2 π$ , the same as the cosine and sine they are built from.
Cotangent has period $π$ , the same as tangent.

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