8.5E Exercises

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The Unit Circle Again

Fill in the blanks. (Try to do this from memory as much as possible.)

$blank unit circle.png$

Answer: If needed, refer back to Section 8.4.

Vocabulary

For the given functions, tell the amplitude, phase shift (if any, and give the direction), and period.

$ f(x) = \sin( x - \pi) $
$ f(x) = 2\cos x $
$ f(x) = \tan (2x) $
$ f(x) = -3 \sin \left( x + \frac{\pi}{2} \right) $
$ f(x) = -\frac{1}{3}\cos (4x) $
$ f(x) = 4\tan \left( \frac{1}{2}x \right) $

Answer: Go to desmos.com and graph both the OG trig function and the transformed one to visually confirm your answers.

Sketching Transformed Trig Functions

Sketch the functions.

$ f(x) = 2\sin x$
$ f(x) = \frac{1}{2} \cos x $
$ f(x) = -3 \sin x $
$ f(x) = \cos(x - \pi) $
$ f(x) = \sin(2x) $
$ f(x) = 2\cos\left( x + \frac{\pi}{2} \right) $
$ f(x) = \sin \left( x - \frac{\pi}{4} \right) $
$ f(x) = \frac{1}{3} \cos \left( \frac{1}{2} x \right) $

Answer: Go to desmos.com and graph the function to visually confirm your answers.

Inverse Trig Function Facts

1. As we saw, the restricted domain we use for sine is $ \left[- \frac{\pi}{2}, \frac{\pi}{2} \right]$. Give the domain and range of arcsine. Go to desmos.com and type in "$y = \sin x \{ \pi/2 \leq x \leq \pi/2\} $" to the first line. Type in "$y = \arcsin x$" to the second line. Describe the relationship you see between the graphs.

2. The restricted domain used for cosine is $ \left[0, \pi \right]$. Give the domain and range of arccosine. Graph both on desmos and describe the relationship.

3. The restricted domain used for tangent is $ \left(- \frac{\pi}{2}, \frac{\pi}{2} \right)$ . Give the domain and range for arctangent. Graph both on desmos and describe the relationship.

Answer

For all three, the graphs should look like reflections across the line $y = x$.

dom: $ [-1,1]$. range: $ \left[- \frac{\pi}{2}, \frac{\pi}{2} \right]$.
dom: $ [-1,1]$. range: $ \left[0, \pi \right]$.
dom: $ (-\infty, \infty) $. range: $ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) $.

Exploring Inverse Reciprocal Trig

1. Go to desmos.com and graph $ f(x) = \operatorname{arccsc} x $. Give the domain and range. By reverse engineering, determine the restricted domain used for $ \csc x$. Graph $g(x) = \csc x$ to see the branch used and compare the graphs.

2. Do the same for $ f(x) = \operatorname{arcsec} x$.

3. Do the same for $ f(x) = \operatorname{arccot} x$.

Answer

dom: $ (-\infty, -1] \cup [1, \infty) $. range: $ \left[- \frac{\pi}{2}, \frac{\pi}{2} \right]$ (this interval is also the restricted domain for cosecant).
dom: $ (-\infty, -1] \cup [1, \infty) $. range: $[0, \pi]$ (this is also the restricted domain for secant).
dom: $ (-\infty, \infty)$. range: $ (0,\pi) $ (this is also the restricted domain for cotangent).

Evaluating Inverse Trig Functions

Evaluate.

$ \arcsin(0)$
$ \arcsin(1) $
$ \arcsin(-1)$
$ \arcsin \left( \frac{\sqrt{3}}{2} \right) $
$ \arcsin \left( -\frac{\sqrt{3}}{2} \right) $
$ \arcsin \left( \frac{1}{\sqrt{2} }\right) $
$ \arccos(0) $
$ \arccos(1)$
$ \arccos(-1)$
$ \arccos \left( \frac{1}{2} \right) $
$ \arccos \left( \frac{1}{\sqrt{2}} \right) $
$ \arccos \left( -\frac{1}{\sqrt{2}} \right) $
$ \arctan(0) $
$ \arctan(1) $
$ \arctan(-1) $
$ \arctan \left( \frac{1}{\sqrt{3}} \right) $

Answer

$0$
$ \pi/2$
$ -\pi/2$
$ \pi/3$
$ -\pi/3$
$ \pi/4$
$ \pi/2$
$0$
$\pi$
$ \pi/3$
$ \pi/4$
$ 3\pi/4$
$ 0$
$ \pi/4$
$ -\pi/4$
$ \pi/6$

Evaluating Inverse Reciprocal Trig Functions

Evaluate $ \operatorname{arcsec} (2) $ by following these steps:

Translate the equation $ \operatorname{arcsec} (2) = \theta$ into an equation involving only secant.
Translate the new equation into an equivalent equation involving only cosine.
Solve the equation for the cosine term.
Find the angle $\theta$ using the unit circle and your knowledge of cosine. Refer back to your exploration of arcsecant to determine which range of angles to report your answer in.

Do the same process to evaluate $ \operatorname{arccsc} \left(\frac{2}{\sqrt{3}} \right)$.

Answer

$ \sec \theta = 2$
$ \frac{1}{\cos \theta} = 2$
$ \cos \theta = \frac{1}{2} $
Cosine is $\frac{1}{2}$ at $\frac{ \pi}{3}$. This is within the allowable range $ [0,\pi] $.

And $\operatorname{arccsc} \left(\frac{2}{\sqrt{3}} \right) = \frac{\pi}{3} $.

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