Graphing Inverse Trig Functions

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Pre-Calculus › Graphing Inverse Trig Functions

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This graph could be the graph of which of the following functions?

Arccot

tan(x)

cot(x)

arctan(x)

arccot(x)

CORRECT

arccsc(x)

Explanation

This is the graph of arccot(x). Think of a few characteristics of the function cot(x): it has a point at (- $\pi$ /2, 0), and vertical asymptotes at x=- $\pi$ , x=0, and x= $\pi$ . Notice that in the graph above, we have the point (0, - $\pi$ /2), which flips the x and y values of the point on cot(x), and we can also see horizontal asymptotes at both y=0 and y= $\pi$ . We don't also have another horizontal asymptote at y=- $\pi$ because we needed to restrict the domain of arccot(x) in order for it to be a function and pass the vertical line test. You can look at the graph of cot(x) below and compare the two and see their similarities:

Cotx

Which quadrant could arccot (−½) fall in?

I and IV

II and IV

CORRECT

III and IV

II and III

Explanation

The cotangent function is negative in quadrants II and IV, so arccot (−½) could fall in either of these quadrants. The below image shows where each function is positive. Any that are not noted are negative. Since cotangent is positive in Quadrants I and III, it is negative in Quadrants II and IV.

Alg2 trig graphik 66

Find the inverse of the function y = sin(x). Then graph both functions.

y = sin-1(x)

Graphs:

y = sin (x)

Sinx

y = sin-1(x)

Arcsinx

CORRECT

y = sin(x)

Graphs:

y = sin(x)

Sinx

y = sin(x)

Sinx

y = cos(x)

Graphs:

y = sin(x)

Sinx

y = cos(x)

Cosine

y = tan(x)

Graphs:

y = sin(x)

Sinx

y = tan(x)

Tan x

y = cos-1(x)

Graphs:

y = sin (x)

Sinx

y = cos-1(x)

Arccos

Explanation

To find the inverse of a function, exchange the values of x and y, then solve for y.

Change y = sin(x) to x = sin(y). Then solve for y:

x = sin(y)

sin-1(x) = sin-1(sin(y))

sin-1(x) = y

Now graph each function:

y = sin(x)

Sinx

y = sin-1(x)

Arcsinx

If you look at the upper graph and restrict the domain to the interval \[- $\pi$ /2, $\pi$ /2\], you can more clearly see the relationship between the two graphs.

Screen shot 2020 06 01 at 1.16.29 pm

To further illustrate the relationship between the two functions, examine the inputs and outputs of the two functions on the interval \[- $\pi$ /2, $\pi$ /2\]:

y = sin(x)

Sintable

y = sin-1(x)

Arcsin table

Evaluate the following expression, assuming that all angles are in Quadrant I:

Screen shot 2020 06 01 at 11.00.58 am

5/12

5/13

12/5

CORRECT

13/5

12/13

Explanation

To solve Screen shot 2020 06 01 at 11.00.58 am , first, let $A=\sin^-^1 \frac{12}{13}$ . Then, . Because this value is positive, we know that this must be in Quadrant I or Quadrant II, but given the instructions, we can assume A is in Quadrant I.

Screen shot 2020 06 01 at 11.23.28 am

Next, recall that $\tan A = \frac{\sin A}{\cos A}$ .

Using the Pythagorean Theorem, you can solve for the following:

x2 + y2 = r2

x2 + 122 = 132

x = 5

Therefore, $\cos A= \frac{5}{13}$ . Now, solve:

Screen shot 2020 06 01 at 11.00.58 am $= \tan A = \frac{\sin A}{\cos A}$ Screen shot 2020 06 01 at 11.46.42 am

Therefore, Screen shot 2020 06 01 at 11.00.58 am Screen shot 2020 06 01 at 11.47.50 am

True or false: None of the inverses of the trigonometric functions are functions.

True

CORRECT

False

Explanation

This is true. Each of the inverse trig functions fails to pass the vertical line test, and is therefore not a function. For this reason, you will also see graphs of the inverse trig functions that restrict the domain such that is can be graphed as a function.

Which quadrant could arcsin (−½) fall in?

I and II

I and III

I and IV

II and III

III and IV

CORRECT

Explanation

The sine function is negative in quadrants III and IV, so arcsin (−½) could fall in either of these quadrants. The below image shows where each function is positive. Any that are not noted are negative. Since sine is positive in Quadrants I and II, it is negative in Quadrants III and IV.

Alg2 trig graphik 66

State the domain and range of $\tan \left ( x \right )$ and $\arctan \left ( x \right )$ .

$\tan (x)$ domain: $(-\infty,\infty)$

$\tan (x)$ range: $(-\infty,\infty)$

$\arctan \left ( x \right )$ domain: $(-\infty,\infty)$

$\arctan \left ( x \right )$ range: $(-\infty,\infty)$

$\tan (x)$ domain: $(-\infty,\infty)$

$\tan (x)$ range: $\left ( -\frac{\pi }{2},\frac{\pi }{2} \right )$

$\arctan \left ( x \right )$ domain: $\left ( -\frac{\pi }{2},\frac{\pi }{2} \right )$

$\arctan \left ( x \right )$ range: $(-\infty,\infty)$

$\tan (x)$ domain: , such that $x\neq \frac{\pi }{2}+n\pi$

$\tan (x)$ range: $(-\infty,\infty)$

$\arctan \left ( x \right )$ domain: $(-\infty,\infty)$

$\arctan \left ( x \right )$ range: $\left ( -\frac{\pi }{2},\frac{\pi }{2} \right )$

CORRECT

$\tan (x)$ domain: $\left ( -\frac{\pi }{2},\frac{\pi }{2} \right )$

$\tan (x)$ range: $(-\infty,\infty)$

$\arctan \left ( x \right )$ domain: $(-\infty,\infty)$

$\arctan \left ( x \right )$ range: , such that $x\neq \frac{\pi }{2}+n\pi$

$\tan (x)$ domain: $\left ( -\frac{\pi }{2},\frac{\pi }{2} \right )$

$\tan (x)$ range: $\left ( -\frac{\pi }{2},\frac{\pi }{2} \right )$

$\arctan \left ( x \right )$ domain: $\left ( -\frac{\pi }{2},\frac{\pi }{2} \right )$

$\arctan \left ( x \right )$ range: $\left ( -\frac{\pi }{2},\frac{\pi }{2} \right )$

Explanation

The two graphs help provide a visual. The domain is all possible x values, and the range is all possible y values. The arctan function exchanges all x and y values of the tangent function. However, in order to continue being a function (by passing the horizontal line test), the arctangent function's range is restricted to $\left ( -\frac{\pi }{2},\frac{\pi }{2} \right )$ . It is not a coincidence that the range of the tangent function is equal to the domain of the arctangent function because of the nature of inverse relationships.

$y=\tan x$

Screen shot 2020 08 21 at 3.28.17 pm

$y=\arctan x$

Screen shot 2020 08 21 at 3.28.22 pm

$\tan (x)$ domain: , such that $x\neq \frac{\pi }{2}+n\pi$

$\tan (x)$ range: $(-\infty,\infty)$

$\arctan \left ( x \right )$ domain: $(-\infty,\infty)$

$\arctan \left ( x \right )$ range: $\left ( -\frac{\pi }{2},\frac{\pi }{2} \right )$

You can see that while the $\tan (x)$ range and the $\arctan \left ( x \right )$ domain match, the $\tan (x)$ domain and the $\arctan \left ( x \right )$ range are not exact matches; this is so that $\arctan \left ( x \right )$ can be a function (by being restricted).

Evaluate the following expression, assuming that all angles are in Quadrant I:

$\sec (\cos ^{-1}1)$

$-\frac{\pi}{2}$

$\frac{\pi}{2}$

CORRECT

Explanation

To solve $\sec (\cos ^{-1}1)$ , first, let $A=\cos ^{-1}1$ . Then, $\cos A=1$ . Because this value is positive, we know that this must be in Quadrant I or Quadrant IV, but given the instructions, we can assume A is in Quadrant I.

Next, recall that $\sec x=\frac{1}{\cos x}$ . Therefore

$\sec (\cos ^{-1}1)=\frac{1}{\cos (\cos ^{-1}1)}$

By definition, $\cos (\cos ^{-1}x)=x$ , assuming that $-1\leq x\leq 1$ , since the cosine function can only output values within this range.

Therefore, $\sec (\cos ^{-1}1)=1$ .

This graph could be which of the following functions?

Arcsec

arcsin(x)

arccos(x)

arctan(x)

arcsec(x)

CORRECT

arccsc(x)

Explanation

This is the graph of arcsec(x). Think of a few points of the function sec(x): (- $\pi$ , -1), (0,1), ( $\pi$ , -1). Now, interchange the x and y values to yield: (-1, - $\pi$ ), (1, 0), (-1, $\pi$ ). Notice that (-1, - $\pi$ ) and (1, 0) both exist on this graph. (-1, $\pi$ ) does not exist on this graph because in order to make this a function, the domain was required to be restricted such that it would pass the vertical line test. You can see how the graph of arcsec(x) relates to the graph of sec(x) by comparing the two:

Secx

State the domain and range of sin(x) and arcsin(x).

sin(x) domain: $(-\infty,\infty)$

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

arcsin(x) domain: $[-\frac{\pi}{2}, \frac{\pi}{2}]$

arcsin(x) range: $(-\infty,\infty)$

sin(x) domain: $(-\infty,\infty)$

sin(x) range: $-1\leq y\leq 1$

arcsin(x) domain: $-1\leq x\leq 1$

arcsin(x) range: $[-\frac{\pi}{2}, \frac{\pi}{2}]$

CORRECT

sin(x) domain: $-1\leq x\leq 1$

sin(x) range: $-1\leq y\leq 1$

arcsin(x) domain: $-1\leq x\leq 1$

arcsin(x) range: $-1\leq y\leq 1$

sin(x) domain: $(-\infty,\infty)$

sin(x) range: $(-\infty,\infty)$

arcsin(x) domain: $(-\infty,\infty)$

arcsin(x) range: $(-\infty,\infty)$

Explanation

The two graphs help provide a visual. The domain is all possible x values, and the range is all possible y values. The arcsin function turns the sine function on its side, but, in order to continue being a function (by passing the horizontal line test), the arcsine function's range is restricted to $[-\frac{\pi}{2}, \frac{\pi}{2}]$ . It is not a coincidence that the range of the sine function is equal to the domain of the arcsin function because of the nature of inverse relationships.