8.3 Trigonometry via Triangles
- Page ID
- 157063
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)By the end of this section, you will be able to:
- understand angle measure in degrees and radians
- identify and find the trigonometric ratios sine, cosine, tangent, secant, cosecant, and cotangent
- use a given trigonometric ratio to find the other ratios at the same angle
The main way that I want you to think about trig functions is going to be using the unit circle in the next section, spoiler alert. However, rather than doggedly memorizing the unit circle, I personally use the special triangles we discussed in Section 8.1, together with the trig ratios we will discuss in this section, to deduce all the information I need about the unit circle from scratch on the fly. Plus, in applications, you will definitely need to remember all this triangle stuff for geometric-style calculus problems, so we'll start here! First, a quick recap of how to measure angles.
Geometry of Angle Measure
"Did a full one-eighty...Crazy..." You already know that turning around to face the exact opposite direction is a \(180^\circ\) turn. Spinning in a full circle to face the exact same way you started is a \(360^\circ\). What we're describing with these numbers is the measure of an angle. We can portray angles using a circle with radius one, by starting with a standard horizontal "starting line" and swinging the angle in a counterclockwise direction:
Angles have two options for their units: degrees (\(^\circ\)) and radians (rad). You might find degrees more natural because high school loves them (God knows why, they're terrible???) but you will pretty much always use radians in future calculus courses (I promise it's better for you, like eating broccoli). These are the important angles you should have on instant recall:
| Angle in Degrees | Angle in Radians | Angle on Circle |
| \(0^\circ\) | \(0\) |  |
| \(30^\circ\) | \( \dfrac{\pi}{6}\) |  |
| \(45^\circ\) | \( \dfrac{\pi}{4}\) |  |
| \(60^\circ\) | \( \dfrac{\pi}{3}\) |  |
| \(90^\circ\) | \( \dfrac{\pi}{2}\) |  |
| \(180^\circ\) | \( \pi\) |  |
| \(270^\circ\) | \( \dfrac{3\pi}{2}\) |  |
| \(360^\circ\) | \( 2\pi\) |  |
Here are all the most commonly used angles on a single unit circle (remember we measure counterclockwise from zero).
It's really best to just memorize these angles, by imagining you're slicing this circle into pizza pieces. You can see that if you cut the top half of the pizza into thirds, the cuts line up with the angles "a third of \(\pi\)" and "two thirds of \(\pi\)," right? For other, weirder angles, if you ever find yourself needing to convert back and forth between degrees and radians, the key is to recognize that \(180^\circ = \pi\) radians. That means you can do a unit conversion with the factors \(\frac{180}{\pi}\) or \(\frac{\pi}{180}\).
To convert from degrees to radians, multiply the angle by \(\dfrac{\pi}{180}\) and simplify.
To convert from radians to degrees, multiply the angle by \( \dfrac{180}{\pi}\) and simplify.
One last thing: You can also talk about negative angles, which are measured in a clockwise direction. For example, the angle \( -\frac{\pi}{3}\) takes you to the same location as \( \frac{5\pi}{3}\). You will practice working with negative angles (and angles bigger than \(360^\circ\)) in the exercises section.
Trigonometric Ratios
Consider a right triangle with a certain acute angle \(\theta\). The trigonometric ratios are certain ratios of the sides of the triangle relative to the chosen angle \(\theta\).
The trigonometric ratios sine, cosine, tangent, cosecant, secant, and cotangent are defined as
\[ \sin \theta = \frac{opp}{hyp} \quad \quad \cos \theta = \frac{adj}{hyp} \quad \quad \tan \theta = \frac{opp}{adj} \notag \]
\[ \csc \theta = \frac{hyp}{opp} \quad \quad \sec \theta = \frac{hyp}{adj} \quad \quad \cot \theta = \frac{adj}{opp} \notag \]
Note: the ratios in the second row are simply the reciprocals of the ratios in the first row:
\[ \csc \theta = \frac{1}{\sin \theta} \quad \quad \sec \theta = \frac{1}{\cos \theta} \quad \quad \cot \theta = \frac{1}{\tan \theta} \notag \]
Also Note: \(\tan \theta = \dfrac{\sin \theta}{\cos\theta}\) and \( \cot \theta = \dfrac{\cos \theta}{\sin \theta}\)! This is very important. Confirm by dividing the side ratios of sine and cosine to see that they turn out to be \(\frac{opp}{adj}\), etc.
You can memorize these ratios with the "SOH-CAH-TOA" mnemonic, which stands for "Sine:Opposite/Hypotenuse, Cosine:Adjacent/Hypotenuse, Tangent:Opposite/Adjacent." The following example demonstrates how to find the values of all these ratios for a particular angle.
Find the trigonometric ratios for \(\theta = \frac{\pi}{6}\).
Solution
First, let's sketch the special right triangle that features the angle \( \frac{\pi}{6} = 30^\circ\), a 30-60-90 triangle. We already know the standard side lengths for this triangle, too.
We identify the opposite side to be \(1\), the adjacent side to be \(\sqrt{3}\), and the hypotenuse to be \(2\). We can now compute the main trig ratios:
\[ \sin \left(\frac{\pi}{6}\right) = \frac{1}{2} \quad \quad \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \quad \quad \tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \notag \]
We can use these ratios to find \( \csc \theta = \frac{1}{\sin \theta}\), \( \sec \theta = \frac{1}{\cos \theta} \), and \( \cot \theta = \frac{1}{\tan \theta}\):
\[ \csc \left(\frac{\pi}{6}\right) = 2 \quad \quad \sec \left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}} \quad \quad \cot \left(\frac{\pi}{6}\right) = \sqrt{3} \notag \]
Find the trigonometric ratios for \( \theta = \frac{\pi}{4} \)
- Answer
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Hint: \(\frac{\pi}{4} = 45^\circ\) so you can use a special triangle.
\[ \sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \quad \quad \cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \quad \quad \tan \left(\frac{\pi}{4}\right) =1 \notag \]\[ \csc \left(\frac{\pi}{4}\right) = \sqrt{2} \quad \quad \sec \left(\frac{\pi}{4}\right) =\sqrt{2} \quad \quad \cot \left(\frac{\pi}{4}\right) = 1 \notag \]
This is great and all for angles less than \( \frac{\pi}{2}\) (\(90^\circ\)), but what about the rest of the unit circle? Let's try sketching our triangles on the plane instead...
I've drawn radii on that circle showing two different angles. Consider the red angle \( \frac{\pi}{3}\). That's \(60^\circ\), so I drop in some dotted lines to create a 30-60-90 right triangle. I also label the side lengths that I know because it's a special triangle. Now I'm ready to compute, say, \(\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \). What about that blue angle, \( \frac{2\pi}{3}\)? Here's the key: drop a vertical line from the endpoint of your radius to the \(x\)-axis. In this case, that's straight downward. Sketch in your triangle and determine the reference angle inside that triangle. The reference angle for \(\frac{2\pi}{3}\) is \(\frac{\pi}{3}\). Now, I label the sides of my special triangle again, but because my adjacent leg is on the negative side of the \(x\)-axis, I label that \(-1\).
At this point, you can use the blue triangle to find any trig ratios you need. For example, \( \sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2}\) and \( \cos \frac{2\pi}{3} = \frac{-1}{2} \). Here are some more examples of reference angles and triangles for different locations on a circle:
Extrapolating Trig Ratios
The last thing I want to do in this section is make sure you can reverse-engineer trig ratios from a single given one. You will be given a value like \( \sin \theta = \frac{a}{b}\) and information about which quadrant the angle lies in. The quadrants of the plane are shown below:
- Sketch a right triangle in the plane in the appropriate quadrant and label \(\theta\). Label two of the sides \(a\) and \(b\), according to the trig ratio given.
- Use the Pythagorean Theorem to find the third side of the triangle.
- Use the triangle to find the other trig ratios as usual.
Find \( \cos \theta, \tan \theta, \csc \theta, \sec \theta,\) and \(\cot \theta\), given that \( \sin \theta = \frac{7}{25}\) and \(\theta\) is in Quadrant II.
Solution
It doesn't matter if our drawing is to scale or correctly proportioned, we just gotta get in the ballpark. I drop a potential angle \(\theta\) into Quadrant II:
I sketch a right triangle and label the known sides, given that \( \sin \theta = \frac{opp}{hyp} = \frac{7}{25}\). The angle with respect to which you determine "opposite" vs "adjacent" in this case is the reference angle, which I've labeled \(\theta'\). I don't know anything about the adjacent side yet, because sine doesn't use him!
To figure out that ? adjacent side, we can use the Pythagorean Theorem.
\[ a^2 + b^2 = c^2 \quad \rightarrow \quad ?^2 + 7^2 = 25^2 \quad \rightarrow \quad ? = \sqrt{576} = 24 \notag \]
Since this triangle is on the negative side of the \(x\)-axis, I label the unknown side with \( -24\). Now I know all three sides of this triangle and I just hone in on that guy to find all the trig ratios.
\[ \sin \theta = \frac{7}{25} \quad \quad \cos \theta = -\frac{24}{25} \quad \quad \tan \theta = -\frac{7}{24} \notag \]
\[ \csc \theta = \frac{25}{7} \quad \quad \sec \theta =- \frac{25}{24} \quad \quad \cot \theta = -\frac{24}{7} \notag \]
My picture is SUPER not to scale, but it doesn't matter, because the numbers themselves will never steer you wrong!
Find all trig ratios for \(\theta\), using the given and the quadrant.
- \( \tan \theta = \sqrt{3} \) in Quadrant I. (Hint: think of \(\sqrt{3}\) as \(\frac{\sqrt{3}}{1}\).)
- \( \cos \theta = -\frac{1}{\sqrt{2}} \) in Quadrant III.
- Answer
-
1. \(\sin \theta = \frac{\sqrt{3}}{2}, \cos \theta = \frac{1}{2}, \tan \theta = \sqrt{3}, \csc \theta = \frac{2}{\sqrt{3}} , \sec \theta = 2, \cot \theta = \frac{1}{\sqrt{3}} \)
2. \(\sin \theta = -\frac{1}{\sqrt{2}}, \cos \theta = -\frac{1}{\sqrt{2}}, \tan \theta = 1, \csc \theta = -\sqrt{2} , \sec \theta = -\sqrt{2}, \cot \theta = 1 \)
Be sure to check out the exercises section, because you will further explore things like coterminal angles and reference angles in there!