Chapter 10
The Unit Circle Definitions
Circling the Triangle: Generalizing Sine and Cosine
The triangle at right suggests that the numerical value of $ ^{} $ is about 1/3.
Your calculator will confirm your intuition, thus reinforcing your faith in both.
Surely SOH CAH TOA is in his heaven, and all’s well in the world. But the Devil
never sleeps! Should the evil one tempt you to enter sin $ 113^{} $ into your calculator, you’ll be confronted with an enigma: $ ^{} $ , an equation that hints at dark mysteries beyond SOH CAH TOA’s realm, for what could $ ^{} $ even mean? It clearly can’t refer to the opposite-to-hypotenuse ratio in a right triangle containing a $ 113^{} $ angle, for no such triangle exists.
This mystery will dissolve once we’ve redefined the trigonometric functions. But can we do that? We used the original definitions to develop the subject, so wouldn’t changing them now invalidate all our previous work? No it will not, because as you’ll see, the new definitions have been carefully designed to preserve all the old results to which you’ve grown accustomed. The new definitions are, in computer terminology, backwards compatible. Here, then, are the most important definitions in all of trigonometry:
The Unit Circle Definitions of sine and cosine.
We define $ $ and $ $ to be, respectively, the x and y coordinates of the point that we reach when we rotate (1, 0) counterclockwise about the origin through an angle of $ $ .
Sine’s unit-circle definition (unlike its SOH CAH TOA definition) tells us what $ ^{} $ means: It is simply the y-coordinate of point P in the figure at right. Clearly, its numerical value is slightly less than 1, so the calculator’s assertion that $ ^{} $ now makes sense. From the same picture, we can also see that $ ^{} $ ( $ P’ $ ’s x-coordinate) is clearly negative, perhaps about -0.4. Your calculator will confirm this: $ ^{} $ .
To verify that the unit circle definitions are compatible with the original SOH CAH TOA definitions, let $ $ be any acute angle, and consider the figure. By the old definition, $ = a/1 = a $ . By the new one, $ $ is $ P’s x $ -coordinate, which is also a. Thus, the two definitions of cosine agree for all acute angles. (So do the two definitions of sine, as you should verify.) Because the “new” sine and cosine extend the original functions’ domains from acute angles to the more general context of all angles, we say that the new definitions generalize sine and cosine. We shall generalize the other trigonometric functions shortly.
Example 1. Find the exact value of $ ^{} $ .
Solution. To use sine’s unit-circle definition, we must rotate $ (1,0) $ counterclockwise about the origin through $ 90^{} $ , as shown in the figure. The y-coordinate of its new location is, by definition, $ ^{} $ . Since, in this case, the point ends up at $ (0,1) $ , we conclude that
\[ \sin90^{\circ}=1. \]
Example 2. Find the exact value of $ ^{} $ .
Solution. To use cosine’s unit-circle definition, we must rotate $ (1,0) $ counterclockwise about the origin through $ 270^{} $ , as shown in the figure. The x-coordinate of its new location is, by definition, $ ^{} $ . Since the point ends up at $ (0,-1) $ , we conclude that
\[ \cos270^{\circ}=0. \]
Important Convention. A negative input for sine or cosine corresponds to a clockwise rotation of $ (1,0) $ .
Example 3. Find the exact value of $ (-90^{}) $ .
Solution. To use cosine’s unit-circle definition, we must rotate $ (1,0) $ clockwise about the origin through $ 90^{} $ , as shown in the figure at right. The x-coordinate of its new location is, by definition, $ (-90^{}) $ . Since the point ends up at $ (0,-1) $ , we conclude that
\[ \cos(-90^{\circ})=0. \]
To ensure that you understand what we’ve done so far, stop here and do a few simple exercises:
Exercises
Find the exact values of….
$ ^{} $
$ ^{} $
$ ^{} $
$ ^{} $
$ ^{} $
sin $ 360^{} $
$ ^{} $
sin 720°
$ ^{} $
$ (-180^{}) $
$ ^{} $
$ -(180^{}) $
sin( $ -720^{} $ )
sin 360000°
$ (-270^{}) $
sin 450°
But Why Generalize?
Be wise: Generalize.
- An old mathematical saw.
The new definitions are not just exercises in generality for generality’s sake. You can grasp their significance by thinking about the graph of $ y = $ , which we’ll draw by considering the figure at right and noting how $ $ varies as $ $ runs through a full rotation: As $ $ runs from $ 0^{} $ to $ 90^{} $ , $ $ increases from 0 to 1; as $ $ runs from $ 90^{} $ to $ 180^{} $ , $ $ decreases from 1 to 0; as $ $ runs from $ 180^{} $ to $ 270^{} $ , $ $ decreases from 0 to -1; and as $ $ runs from $ 270^{} $ to $ 360^{} $ , $ $ rises from -1 to 0. Putting this all together, we obtain the following graph.
But this is just the beginning. Taking a second lap around the unit circle, as $ $ runs from $ 360^{} $ to $ 720^{} $ , $ $ goes through the same cycle of values it went through during the first lap. The same, of course, holds true in the third lap, the fourth, and so on indefinitely. Throwing in negative values of $ $ for good measure, the graph of sine reveals its true character at last.
This is the famous sine wave, whose endlessly repeating pattern appears not only in mathematical models of waves (water, sound, light), but of periodic phenomena of all sorts, from the revolutions of planets to the vibrations of atoms. By generalizing the trigonometric functions, we thus open up new worlds of applications that would have been unthinkable in the restricted context of right triangles.
Exact Values of Sine and Cosine at Special Angles
God sends special angles into our lives.
- A roadside sign in southern Washington
You’ve already learned the exact values of sine and cosine at the three special angles $ 30^{} $ , $ 60^{} $ , and $ 45^{} $ . We can now find the exact values of sine and cosine at all integer multiples of these three special angles. The method is simple: Draw a picture, and use it to relate the coordinate that you want to something in the first quadrant – the acute-angled land we know so well.
Example 1. Find the exact value of $ (120^{}) $ .
$ (^{}, ^{}) $
Solution. Since $ 120^{} $ falls $ 60^{} $ short of a half-turn, we want the y-coordinate of point P. Clearly, points P and Q have the same y-coordinate, so $ 120^{} $ and $ 60^{} $ have the same sine. Thus,
\[ \sin(120^{\circ})=\sin(60^{\circ})=\frac{\sqrt{3}}{2}. \]
Example 2. Find the exact value of $ (135^{}) $ .
Solution. Since $ 135^{} $ falls $ 45^{} $ short of a half-turn, we must find point $ P’ $ ’s x-coordinate. Clearly, points P and Q have equal but opposite x-coordinates (same magnitude, opposite sign). Hence, $ 135^{} $ and $ 45^{} $ have equal but opposite cosines. Thus,
\[ \cos(135^{\circ})=-\cos45^{\circ}=-\frac{\sqrt{2}}{2}. \]
Example 3. Find the exact value of $ (330^{}) $ .
Solution. Since $ 330^{} $ falls $ 30^{} $ short of a full rotation, we must find point $ P’ $ ’s x-coordinate. Points P and Q have the same x-coordinate, so $ 330^{} $ and $ 30^{} $ have the same sine. Thus,
\[ \cos(330^{\circ})=\cos30^{\circ}=\frac{\sqrt{3}}{2}. \]
Example 4. Find the exact value of $ (210^{}) $ .
Solution. $ 210^{} $ falls $ 30^{} $ beyond a half-turn, so we want $ P’ $ ’s y-coordinate. Clearly, points P and Q have equal but opposite y-coordinates. Hence, $ 210^{} $ and $ 30^{} $ have equal but opposite sines. Thus,
\[ \sin210^{\circ}=-\sin30^{\circ}=-\frac{1}{2}. \]
Exercises
Find the exact values of…
Circling the Triangle II: Generalizing the Tangent Function
He experienced both bliss and horror in contemplating the way an inclined line, rotating spokelike, slid upwards along another, vertical one… The vertical one was infinite, like all lines, and the inclined one, also infinite, sliding along it and rising ever higher… was doomed to eternal motion, for it was impossible for it to slip off, and the point of their intersection, together with his soul, glided upwards along an endless path. But with the aid of a ruler he forced them to unlock: he simply redrew them, parallel to one another, and this gave him the feeling that out there, in infinity, where he had forced the inclined line to jump off, an unthinkable catastrophe had taken place, an inexplicable miracle, and he lingered long in those heavens where earthly lines go out of their mind.
- Vladimir Nabokov, The Defense (Chapter 2).
Having generalized sine and cosine, we will now do as much for tangent, whose unit-circle definition explains this function’s name, since it involves a line that is literally tangent to the unit circle.
The Unit Circle Definition of Tangent.
Imagine a line that initially coincides with the x-axis.
Rotate it about the origin through an angle of $ $ (counterclockwise).
It intersects the fixed line x = 1 at some point.
We define $ $ to be the y-coordinate of that point.
We can easily verify that this new definition of tangent agrees with the old one for all acute angles. Consider the figure above. By the old SOH CAH TOA definition, $ $ is the ratio (in the right triangle) of the vertical to the horizontal leg. But the horizontal leg’s length is 1, so $ $ is simply the vertical leg’s length. By the new unit-circle definition, $ $ is the marked point’s y-coordinate… which is obviously the length of the triangle’s vertical leg. Hence, the two definitions of tangent agree for all acute angles. However, the unit circle definition also applies to non-acute angles, where SOH CAH TOA fears to tread.
Example. Find the exact value of $ (150^{}) $ .
Solution. $ 150^{} $ falls $ 30^{} $ short of a half-turn, so we want $ P^{}s $ y-coordinate, which is clearly equal but opposite to $ Q^{}s $ . The tangents of $ 150^{} $ and $ 30^{} $ are therefore equal but opposite. Thus,
\[ \tan150^{\circ}=-\tan30^{\circ}=-\frac{1}{\sqrt{3}}.\quad\blacklozenge \]
Exercises
Find the exact values of the tangents of: 210°, 240°, 330°, 135°, 120°, 0°, 180°, 300°.
Strictly speaking, tan is undefined at $ 90^{} $ and $ 270^{} $ . Informally, we say that tan is infinite at those angles. Why?
Explain why it makes some intuitive sense to say that $ 1/= 0 $ . (We’ll use this idea in the next section).
State the ranges of the sine, cosine, and tangent functions.
Circling the Triangle III: Last Generalizations
In one word… one must generalize; this is a necessity that imposes itself on the most circumspect observer.
- Henri Poincaré, The Value of Science, Chapter 5
Generalizing the reciprocal trigonometric functions is trivial:
Definitions. For all values of $ $ , we define:
\[ \mathbf{s e c}\theta=\frac{1}{\cos\theta}, \]
\[ \mathbf{c s c}\theta=\frac{1}{\sin\theta}, \]
\[ \mathbf{c o t}\theta=\frac{1}{\tan\theta}. \]
Note: When \(\tan \theta\) is infinite, we define \(\cot \theta\) to be 0. (See exercises 30 and 31 above.)
For example, since $ ^{} = -1/ $ (the previous page’s example), it follows that $ ^{} = - $ .
Having generalized all of the trigonometric functions, it is now incumbent upon us to prove that the identities we proved in the previous chapter still hold in the trigonometric functions’ enlarged domains. (Our original proofs of those identities used right triangles, so they are valid only for acute angles.)
Claim 1. For all values of $ $ at which the following expressions are defined, $ = $ .
Proof. We’ve already proved that this holds for acute angles. Next,
suppose $ $ lies in the second quadrant (as at right).
Drop a perpendicular PF from P to the x-axis, thus producing similar right triangles, $ POF $ and $ TOE $ . Corresponding ratios of similar triangles are equal, so we have
\[ \frac{ET}{OE}=\frac{FP}{FO}. \]
We’ll now express each of these four lengths in terms of $ $ .
From the figure, observe that $ $ is negative, while length ET is (like all lengths) positive. Consequently, $ ET = -$ . Similarly, $ $ is negative, but FO is positive, so $ FO = -$ . The remaining legs are easy: $ FP = $ , and OE = 1. Substituting these boldface expressions into the proportion above and simplifying, we obtain
\[ \tan\theta=\frac{\sin\theta}{\cos\theta}, \]
thus proving that the identity holds for all angles in the second quadrant. Similar arguments (with similar triangles) show that it holds in the third and fourth quadrants, too, as you should verify.
Next, we’ll demonstrate that the cofunction identities hold for all angles. I’ll provide some, but not all, of the details – enough for you to fill in the rest if you so desire.
Claim 2. Sine and cosine are cofunctions. That is, for all values of $ $ ,
\[ \sin(90^{\circ}-\theta)=\cos\theta\qquad and\qquad\cos(90^{\circ}-\theta)=\sin\theta. \]
Proof. We’ve already proved this for acute (i.e. quadrant 1) angles, so now let’s prove that these cofunction identities still hold when lies in quadrant 2, as at right. We’ll do so as follows:
$$
( , )
$$
First, let’s locate $ 90^{}-$ on the unit circle. Subtracting a $ 2^{nd} $ -quadrant $ $ from $ 90^{} $ yields a negative angle (between $ -90^{} $ and $ 0^{} $ ), so rotating (1,0) through $ 90^{}-$ puts it in quadrant 4 (at point Q).
Next, we’ll get some congruent right triangles into the picture by dropping perpendiculars from P and Q to the axes. These congruent triangles tell us that $ BQ = AP $ . But carefully considering the figure shows that $ BQ = -(90^{} - ) $ and $ AP = -$ . Inserting these
expressions into BQ = AP and multiplying both sides by −1, we obtain sin(90° − θ) = cos θ.
A nearly identical argument that begins by observing the equality OB = OA leads, as you should verify, to the conclusion that $ (90^{} - ) = $ .
Arguments very similar to these will prove the identities when $ $ is in quadrants 3 or 4.
Thanks to the work we’ve done already, generalizing the other co-function identities is very easy.
Claim 3. Tangent and cotangent are cofunctions. That is, for all relevant values of $ $ ,
\[ \mathbf{t a n}(90^{\circ}-\theta)=\mathbf{c o t}\theta\qquad\mathrm{a n d}\qquad\mathbf{c o t}(90^{\circ}-\theta)=\mathbf{t a n}\theta. \]
Proof. $ (90{}-)=\frac{(90{}-)}{(90^{}-)}====. $
Those equals are justified (in order) by: Claim 1, Claim 2, algebra, Claim 1, cotangent’s definition. This establishes one identity. The other identity can be proved similarly, as you should verify.
Claim 4. Secant and cosecant are cofunctions. That is, for all relevant values of $ $ ,
\[ \mathbf{sec}(90^\circ-\theta)=\csc\theta\qquad and\qquad\mathbf{csc}(90^\circ-\theta)=\sec\theta. \]
Proof. $ (90{}-)=\frac{1}{(90{}-)}==. $
The equals are justified, respectively, by secant’s definition, Claim 2, and cosecant’s definition. This establishes one identity. The other identity can be proved similarly, as you should verify.
Exercises
- Find the exact values of the following:
\[ cot210^{\circ},\quad sec240^{\circ},\quad csc330^{\circ},\quad sec135^{\circ},\quad csc120^{\circ},\quad sec0^{\circ},\quad csc765^{\circ},\quad cot300^{\circ},\quad cot90^{\circ}. \]
- Which of these numbers are in secant’s range?: -5, $ 2/3 $ , $ -e/$ , $ 10^{10} $ , $ /e $ .
$ (, ) $
- In the figure, we know P’s coordinates in terms of $ $ , while O’s are fixed at $ (0,0) $ . Consequently, we should be able to express segment OP’s slope in terms of $ $ . Do so. [A-ha! Now we have another nice new way to “see” the value of $ $ on the unit circle: It’s simply line OP’s slope! With this new insight, redo exercises 29 & 30. Which of the two unit-circle interpretations of $ $ do you prefer to use?]
The Pythagorean Identity, Even and Odd Functions
Let’s begin with the best-known of all trigonometric identities, often called “the Pythagorean Identity”.
Claim 1. (The Pythagorean Trig Identity) For all values of $ $ , the following relationship holds:
\[ \cos^{2}\theta+\sin^{2}\theta=1.^{*} \]
Proof. For all values of $ $ , the definitions of sine and cosine ensure that the point $ (, ) $ lies on the unit circle. Its coordinates therefore satisfy the unit circle’s equation, $ x^{2} + y^{2} = 1 $ . That is, $ ^{2}+ ^{2}= 1 $ for all values of $ $ , as claimed.
The “Pythagorean” nature of the identity is explained by the figure at right. You’ll encounter the Pythagorean identity often in the future – and not just in this course. Because it relates $ $ and $ $ , it lets us replace expressions involving $ $ with expressions involving $ $ (or vice versa) when doing so is convenient – which is surprisingly often.
An even function is one that always sends numbers and their negatives to precisely the same value. (In symbols, f is even if $ f(-x) = f(x) $ for all x.) Examples: $ f(x) = x^{2} $ , $ g(x) = x^{4} $ , $ h(x) = x^{6} $ , and…
Claim 2. Cosine is an even function. That is, for all values of $ $ ,
\[ \mathbf{c o s}(-\boldsymbol{\theta})=\mathbf{c o s}\boldsymbol{\theta}. \]
Proof. This should be obvious if you understand cosine’s unit circle definition. The x-coordinates of P and Q will clearly be equal no matter what $ $ is (even if $ $ lies outside of the first quadrant). And these equal x-coordinates are, by the unit-circle definition of cosine, $ $ and $ (-) $ respectively.
Since cosine is even, we can quickly note, for example, that cos(−60°) = cos 60° = 1/2.
An odd function is one that always sends numbers and their negatives to equal but opposite values. (In symbols, f is odd if $ f(-x) = -f(x) $ for all x.) Examples: $ f(x) = x^{3} $ , $ g(x) = x^{5} $ , $ h(x) = x^{7} $ , and…
Claim 3. Sine and tangent are odd functions. That is, for all values of $ $ ,
\[ \mathbf{s i n}(-\theta)=-\mathbf{s i n}\theta\qquad\mathrm{a n d}\qquad\mathbf{t a n}(-\theta)=\mathbf{t a n}\theta. \]
Proof. The y-coordinates of P and Q will clearly be equal but opposite no matter what $ $ is (even if $ $ lies outside of the first quadrant). Thus, by the unit circle definition of sine, we have $ (-) = -$ for all $ $ .
The same is true of R and S. Thus, by unit-circle definition of tangent, $ (-) = $ for all $ $ .
Since sine is odd, we have, for example, $ (-60^{}) = -^{} = -/2 $ .
The reciprocal trig functions inherit the evenness or oddity of their “parents”. For example, secant is even for the following reason: For any number $ $ , we have $ (-)=1/(-)=1/=$ . (The equalities are justified by, in order, secant’s definition, cosine’s evenness, and secant’s definition.) Trigonometry is unusual in that most of its main functions are even or odd. Most non-trigonometric functions are neither even nor odd.
Exercises
Give an example of a function that is neither even nor odd, and prove that this is so.
Is cosecant even, odd, or neither? Prove it.
Is cotangent even, odd, or neither? Prove it.
Dividing both sides of the Pythagorean identity by $ ^{2}$ , we obtain a new identity. What is it?
Dividing both sides of the Pythagorean identity by $ ^{2}$ , we obtain a new identity. What is it?
Discover an identity relating $ (180^{}-) $ and $ $ by thinking about the picture of the unit circle at right. This is a handy identity, but you need not memorize it, since any time you need it, you can recall it by drawing the appropriate picture and thinking for a moment.
\[ \left(\cos(180^{\circ}-\theta\right) \]
\[ \sin(180^{\circ}-\theta) \]
Now come up with an identity for $ (180^{}-) $ . This identity, too, is often handy, but once again, you need not memorize it; whenever you need it, just draw the appropriate picture and recover it.
By drawing appropriate pictures of the unit circle and using identities you already know, find identities for the following expressions. As in the two previous exercises, you need not memorize the identities you discover; instead, you should be able to produce them when you need them by drawing a quick sketch and thinking.
\[ \sin(\theta+90^{\circ}) \]
\[ \cos(\theta+90^{\circ}) \]
\[ c)\tan(\theta+90^{\circ}) \]
\[ \sin(\theta-90^{\circ})^{*} \]
\[ \cos(\theta-90^{\circ}) \]
\[ \cos(\theta+180^{\circ}) \]
\[ cos(\theta-180^{\circ}) \]
\[ \sin(\theta+180^{\circ}) \]
\[ \tan(180^{\circ}+\theta) \]
\[ \tan(\theta-180^{\circ}) \]
- Use identities to simplify the following expressions:
$ $ b) $ $ c) $ ()^{2}+1 $ [Hint: Exercise 46.]
$ $ e) $ $ f) $ (180{}+)(180{}-)-(180{}-)(+90{}) $
Every even function’s graph is symmetric about the y-axis. By thinking about the definition of an even function, explain why this is so.
The graphs of all odd functions display a peculiar sort of symmetry. Discover it and describe it.
I claim that if an odd function is defined at zero, then its graph must pass through the origin.
Either use the definition of an odd function to prove my claim or provide a counterexample that disproves it.
- There’s only one function whose domain is all real numbers that is both even and odd. What is it?
[Hint: Think geometrically.]