8.4E Exercises

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The Unit Circle

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

$blank unit circle.png$

Answer: AFTER you try as much as possible, you can look back to the previous section for hints. Practice using Quadrant I to deduce the rest of the blanks!

Using the Unit Circle

Use the unit circle to evaluate the trig functions.

$ \sin (0)$
$ \cos \left( \frac{3\pi}{2} \right) $
$ \tan(0)$
$ \sin \left( \frac{\pi}{6} \right) $
$ \cos \left( \frac{2\pi}{3} \right) $
$ \tan \left( \frac{\pi}{4} \right) $
$ \sec (\pi) $
$ \csc \left( \frac{\pi}{4} \right) $
$ \cot \left( \frac{5\pi}{6} \right) $

Answer

$0$
$0$
$0$
$ \frac{1}{2}$
$ -\frac{1}{2}$
$ 1$
$ -1$
$ \sqrt{2}$
$ -\sqrt{3}$

Sketching Basic Trig Functions

From memory and using your knowledge of the unit circle, sketch the graphs of $ \sin x, \cos x, $ and $ \tan x $, labeling intercepts and any asymptotes.

Answer: You can check by comparing to the previous section.

Exploring Cosecant

1. How is cosecant related to sine?

2. At which $x$ values is $\sin x$ equal to 0? What can you say about cosecant at those locations?

3. At which $x$ values is $ \sin x$ equal to $ \csc x$?

4. Look at the graph of sine below and imagine approaching $x = 0$ from the right. (Aka, your angle value $x$ is positive but getting smaller.) What are the function values ($y$-values) doing as you approach? What do you expect the function values of cosecant to do as $x$ approaches $0$ from the right, as a result? Check your hypothesis by looking at the graph of $\csc x$ in the Answer.

$sine.png$

5. Go to desmos.com and graph both sine and cosecant together on the same graph.

Answer

$ \csc x = \frac{1}{\sin x} $
$ \sin x = 0$ if $x = ..., -2\pi, -\pi, 0, \pi, 2\pi, ...$ etc. Since we can't divide by 0, cosecant is undefined at these locations.
Whenever $ \sin x = \pm 1$, $\csc x = \frac{1}{\pm 1} = \pm 1$. That happens at $x = -\pi/2, \pi/2, 3\pi/2, 5\pi/2, ...$ etc.
Coming into $x = 0$ from the right, the graph of sine is decreasing toward the origin. Aka the function values are positive but getting super tiny. If $ \sin x = \frac{1}{1000}$, for example, $ \csc x = 1000$, so when sine gets tiny, cosecant blows up. Indeed, we see that approaching $x = 0$ from the right, the graph below blows upward along the vertical asymptote! Similar reasoning explains the behavior near the other asymptotes as well.

$csc.png$

Exploring Secant

1. How is secant related to cosine?

2. At which $x$ values is $\cos x$ equal to 0? What can you say about secant at those locations?

3. At which $x$ values is $ \cos x$ equal to $ \sec x$?

4. Look at the graph of cosine below and imagine approaching $x = \frac{\pi}{2}$ from the left. What are the function values ($y$-values) doing as you approach? What do you expect the function values of secant to do as $x$ approaches $\frac{\pi}{2}$ from the left, as a result? Check your hypothesis by looking at the graph of $\sec x$ in the Answer.

$cosine.png$

5. Go to desmos.com and graph both cosine and secant together on the same graph.

Answer

$\sec x = \frac{1}{\cos x}$
$ \cos x = 0$ when $x = ... -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, ...$ etc. Then secant is undefined at those inputs.
Anytime $\cos x = \pm 1$. This happens at $x = ... -\pi, 0, \pi, 2\pi, ...$ etc.
Coming in to $ x = \pi/2$ from the left, the $y$-values cosine gives us are positive but getting smaller. We expect the values of secant to get larger. Indeed, in the graph we see secant blow up along the left side of the vertical asymptote $x = \pi/2$.

$sec.png$

Exploring Cotangent

1. What is the relationship between cotangent and tangent? What is the relationship between cotangent and sine and cosine?

2. At what $x$-values is $\tan x$ equal to zero? What can you say about $\cot x$ at those locations?

3. Where is $ \cot x $ equal to zero?

4. If $ \tan x $ is a large positive number at some input value $x$, what do you expect about the value of $ \cot x$ there? If $\cot x$ crosses the $x$-axis at an $x$-value, what do you know about tangent at that location? Hint: look back at your relationship between tan and cotan.

5. Go to desmos.com and graph both tangent and cotangent together on the same graph.

Answer

$ \cot x = \frac{1}{\tan x} = \frac{ \cos x }{\sin x}$
$\tan x = 0$ anytime $\sin x = 0$, which we've already seen is at $x = ..., -2\pi, -\pi, 0, \pi, 2\pi, ...$ etc. At those $x$-values, cotan is undefined.
$ \cot x = 0$ anytime $ \cos x = 0$, which we've already seen is at $x = ... -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, ...$ etc.
We expect $ \cot x$ to be a very small (but positive) number like a teeny fraction. If $ \cot x$ has a root at $x$, then $ \tan x = \frac{1}{\cot x} $ will be undefined!

Law of Cosines

We're all familiar by now with the Pythagorean Theorem, but it only applies to right triangles. Turns out, there is a generalization of the Pythagorean Theorem called the Law of Cosines. For the triangle below,

$lawofcosines.png$

the Law of Cosines says that

\[ c^2 = a^2 + b^2 -2ab \cos \theta, \notag \]

where $\theta$ is the angle opposite side $c$. Being a "generalization" means that this is a broader overarching rule, of which the Pythagorean Theorem is supposed to be a special case. Explain why the Pythagorean Theorem can be derived as a special case of the Law of Cosines.

Answer: Hint: What happens if $ \theta$ is a right angle? What do you know about the function cosine?

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