In Section circularmotion, we introduced circular motion and derived a formula which describes the linear velocity of an object moving on a circular path at a constant angular velocity. One of the goals of this section is describe the position of such an object. To that end, consider an angle \(\theta \) in standard position and let \(P\) denote the point where the terminal side of \(\theta \) intersects the Unit Circle, as diagrammed below.

By associating the point \(P\) with the angle \(\theta \), we are assigning a position on the Unit Circle to the angle \(\theta \). Since for each angle \(\theta \), the terminal side of \(\theta \), when graphed in standard position, intersects The Unit Circle only once, the mapping of \(\theta \) to \(P\) is a function. Since there is only one way to describe a point using rectangular coordinates, the mappings of \(\theta \) to each of the \(x\) and \(y\) coordinates of \(P\) are also functions. We give these functions names in the following definition.

You may have already seen definitions for the sine and cosine of an (acute) angle in terms of ratios of sides of a right triangle. While not incorrect, defining sine and cosine using right triangles limits the angles we can study to acute angles only. Definition sinecosineunitcircledefn, on the other hand, applies to all angles. Since these functions are defined in terms of points on the Unit Circle, they are called circular functions. Rest assured, Definition sinecosineunitcircledefn specializes to Definition righttrianglesinecosinetangent when \(\theta \) is an acute angle. We will see instances of this fact in the next example.

A few remarks are in order. First, after having re-used some of our work from Section AppRightTrig in a few specific instances, we can reconcile Definition sinecosineunitcircledefn with Definition righttrianglesinecosinetangent in the case \(\theta \) is an acute angle. We situate \(\theta \) in a right triangle with hypotenuse length \(1\), adjacent side length ‘\(x\),’ and the opposite side length ‘\(y\)’ as seen below on the left. Placing the vertex of \(\theta \) at the origin and the adjacent side of \(\theta \) along the \(x\)-axis as seen below on the right effectively puts \(\theta \) in standard position with \(\theta \)’s adjacent side as the initial side of \(\theta \) and the hypotenuse as the terminal side of \(\theta \). Since the hypotenuse of the triangle has length \(1\), we know the point \(P(x,y)\) is on the Unit Circle.

Definition righttrianglesinecosinetangent gives \(\cos (\theta ) = \frac {x}{1} = x\) and \(\sin (\theta ) = \frac {y}{1} = y\) which exactly matches Definition sinecosineunitcircledefn. Hence, in the case of acute angles, the two definitions agree. In other words, the values of the trigonometric ratios of acute angles are the same as the corresponding circular function values.

A second important take-away from Example cosinesineviaunitcircle is use of symmetry in number refangleintro. Indeed, we found the sine and cosine of \(\frac {5\pi }{6}\) using the (acute) angle \(\frac {\pi }{6}\) ‘for reference.’ Since the Unit Circle is rife with symmetry, we would like to generalize this concept and exploit symmetry whenever possible. To that end, we introduce the notion of reference angle.

In general, for a non-quadrantal angle \(\theta \), the reference angle for \(\theta \) (which we’ll usually denote \(\alpha \)) is the acute angle made between the terminal side of \(\theta \) and the \(x\)-axis. If \(\theta \) is a Quadrant I or IV angle, \(\alpha \) is the angle between the terminal side of \(\theta \) and the positive \(x\)-axis:

If \(\theta \) is a Quadrant II or III angle, \(\alpha \) is the angle between the terminal side of \(\theta \) and the negative \(x\)-axis:

If we let \(P\) denote the point \((\cos (\theta ), \sin (\theta ))\), then \(P\) lies on the Unit Circle. Since the Unit Circle possesses symmetry with respect to the \(x\)-axis, \(y\)-axis and origin, regardless of where the terminal side of \(\theta \) lies, there is a point \(Q\) symmetric with \(P\) which determines \(\theta \)’s reference angle, \(\alpha \). The only difference between the points \(P\) and \(Q\) are the signs of their coordinates, \(\pm \). Hence, we have the following:

In light of Theorem refanglethm, it pays to know the sine and cosine values for certain common Quadrant I angles as well as to keep in mind the signs of the coordinates of points in the given quadrants.

\[ \begin{array}{|c|c||c|c|} \hline \theta (\mbox {degrees}) & \theta (\mbox {radians}) & \cos (\theta ) & \sin (\theta ) \\ \hline 0^{\circ } & 0 & 1 & 0 \\ \hline 30^{\circ } & \frac {\pi }{6} & \frac {\sqrt {3}}{2} & \frac {1}{2} \\ \hline 45^{\circ } & \frac {\pi }{4} & \frac {\sqrt {2}}{2} & \frac {\sqrt {2}}{2} \\ \hline 60^{\circ } & \frac {\pi }{3} & \frac {1}{2} & \frac {\sqrt {3}}{2} \\ \hline 90^{\circ } & \frac {\pi }{2} & 0 & 1 \\ \hline \end{array} \]

A couple of remarks are in order. First off, the reader may have noticed that when expressed in radian measure, the reference angle for a non-quadrantal angle is easy to spot. Reduced fraction multiples of \(\pi \) with a denominator of \(6\) have \(\frac {\pi }{6}\) as a reference angle, those with a denominator of \(4\) have \(\frac {\pi }{4}\) as their reference angle, and those with a denominator of \(3\) have \(\frac {\pi }{3}\) as their reference angle.

Also note in number coterminalsamesinecosineexample above, the angles \(\frac {\pi }{3}\) and \(\frac {7\pi }{3}\) are coterminal. As a result, have the same values for sine and cosine. It turns out that we can characterize coterminal angles in this manner, as stated below.

Recall the phraseology ‘if and only if’ means there are two things to argue in Theorem coterminalsamcosinesinethm: first, if \(\alpha \) and \(\beta \) are co-terminal, then \(\cos (\alpha ) = \cos (\beta )\) and \(\sin (\alpha ) = \sin (\beta )\). This is immediate since coterminal share terminal sides, and, in particular, the (unique) point on the Unit Circle shared by said terminal side. Second, we need to argue that if \(\cos (\alpha ) = \cos (\beta )\) and \(\sin (\alpha ) = \sin (\beta )\), then \(\alpha \) and \(\beta \) are coterminal.

To prove this second claim, note that when an angle is drawn in standard position, the terminal side of the angle is the ray that starts at the origin and is completely determined by any other point on the terminal side. If \(\cos (\alpha ) = \cos (\beta )\) and \(\sin (\alpha ) = \sin (\beta )\), then their terminal sides share a point on the Unit Circle, namely \((\cos (\alpha ), \sin (\alpha )) = (\cos (\beta ), \sin (\beta ))\). Hence, \(\alpha \) and \(\beta \) are coterminal.

Combining the Reference Angle Theorem along with our knowledge of the cosine and sine values of common angles, we can produce the figure of the Unit Circle below. We recommend committing it to memory.

Our next example uses The Reference Angle Theorem in a slightly more sophisticated context.

A couple of remarks about Example advancedrefangleex are in order. First, we note the right triangle we used to find \(\sin (\alpha )\) is a scaled 5-12-13 triangle. Recognizing this Pythagorean Triple may have simplified our workflow. Along the same lines, since, the Unit Circle, by definition, is described by the equation \(x^2+y^2 = 1\), we could substitute \(x = \frac {5}{13}\) in order to find \(y\). We leave it to the reader to show we get the exact same answer regardless of the approach used.

Our next example turns the tables and makes good use of the Unit Circle values as well as Theorem coterminalsamcosinesinethm in a different way: instead of giving information about the angle and asking for sine or cosine values, we are given sine or cosine values and asked to produce the corresponding angles. In other words, we solve some rudimentary equations involving sine and cosine.

One of the key items to take from Example solveforangle is that, in general, solutions to trigonometric equations consist of infinitely many answers. To get a feel for these answers, the reader is encouraged to follow our mantra from Chapter SequencesandtheBinomialTheorem - that is, ‘When in doubt, write it out!’ This is especially important when checking answers to the exercises.

For example, another Quadrant IV solution to \(\sin (\theta ) = -\frac {1}{2}\) is \(\theta = -\frac {\pi }{6}\). Hence, the family of Quadrant IV answers to number sineisnegativehalf above could just have easily been written \(\theta = -\frac {\pi }{6} + 2\pi k\) for integers \(k\). While on the surface, this family may look different than the stated solution of \(\theta = \frac {11\pi }{6} + 2\pi k\) for integers \(k\), we leave it to the reader to show they represent the same list of angles.

It is also worth noting that when asked to solve equations in algebra, we are usually looking for real number solutions. Since we can identify radians with real numbers (see Section RadianMeasure), we are able to regard the inputs to the sine and cosine functions as real numbers by identifying any real number \(t\) with an oriented angle \(\theta \) measuring \(\theta = t\) radians. That is, for each real number \(t\), we associate an oriented arc \(t\) units in length with initial point \((1,0)\) and endpoint \(P(\cos (t), \sin (t))\).

In practice this means in expressions like ‘\(\cos (\pi )\)’ and ‘\(\sin (2)\),’ the inputs can be thought of as either angles in radian measure or real numbers, whichever is more convenient.

Suppose, as in the Exercises, we are asked to find all real number solutions to the equation such as \(\sin (t) = -\frac {1}{2}\). The discussion above allows us to find the real number solutions to this equation by thinking in angles. Indeed, we would solve this equation in the exact way we solved \(\sin (\theta ) = -\frac {1}{2}\) in Example solveforangle number sineisnegativehalf. Our solution is only cosmetically different in that the variable used is \(t\) rather than \(\theta \): \(t = \frac {7\pi }{6} + 2\pi k\) or \(t = \frac {11\pi }{6} + 2\pi k\) for integers, \(k\).

We will study the sine and cosine functions in greater detail in Section GraphsofSineandCosine. Until then, keep in mind that any properties of the sine and cosine functions developed in the following sections which regard them as functions of angles in radian measure apply equally well if the inputs are regarded as real numbers.

1 Beyond the Unit Circle

In Definition 1, we define the sine and cosine functions using the Unit Circle, \(x^2+y^2=1\). It turns out that we can use any circle centered at the origin to determine the sine and cosine values of angles. To show this, we essentially recycle the same similarity arguments used in Section ?? to show the trigonometric ratios described in Definition ?? are independent of the choice of right triangle used.

Consider for the moment the acute angle \(\theta \) drawn below in standard position. Let \(Q(x,y)\) be the point on the terminal side of \(\theta \) which lies on the circle \(x^2+y^2 = r^2\), and let \(P(x',y')\) be the point on the terminal side of \(\theta \) which lies on the Unit Circle. Now consider dropping perpendiculars from \(P\) and \(Q\) to create two right triangles, \(\Delta OPA\) and \(\Delta OQB\). These triangles are similar, thus it follows that \(\frac {x}{x'} = \frac {r}{1} = r\), so \(x = r x'\) and, similarly, we find \(y = r y'\). Since, by definition, \(x' = \cos (\theta )\) and \(y' = \sin (\theta )\), we get the coordinates of \(Q\) to be \(x = r \cos (\theta )\) and \(y = r \sin (\theta )\). By reflecting these points through the \(x\)-axis, \(y\)-axis and origin, we obtain the result for all non-quadrantal angles \(\theta \), and we leave it to the reader to verify these formulas hold for the quadrantal angles as well.

Not only can we describe the coordinates of \(Q\) in terms of \(\cos (\theta )\) and \(\sin (\theta )\) but since the radius of the circle is \(r = \sqrt {x^2 + y^2}\), we can also express \(\cos (\theta )\) and \(\sin (\theta )\) in terms of the coordinates of \(Q\). These results are summarized in the following theorem.

Note that in the case of the Unit Circle we have \(r = \sqrt {x^2+y^2} = 1\), so Theorem 3 reduces to our definitions of \(\cos (\theta )\) and \(\sin (\theta )\) in Definition 1. Our next example makes good use of Theorem 3.

Theorem 3 gives us what we need to ‘circle back’ to the question posed at the the beginning of the section: how to describe the position of an object traveling in a circular path of radius \(r\) with constant angular velocity \(\omega \). Suppose that at time \(t\), the object has swept out an angle measuring \(\theta \) radians. If we assume that the object is at the point \((r,0)\) when \(t=0\), the angle \(\theta \) is in standard position. By definition, \(\omega = \frac {\theta }{t}\) which we rewrite as \(\theta = \omega t\). According to Theorem 3, the location of the object \(Q(x,y)\) on the circle is found using the equations \(x = r \cos (\theta ) = r \cos (\omega t)\) and \(y = r \sin (\theta ) = r \sin (\omega t)\). Hence, at time \(t\), the object is at the point \((r \cos (\omega t), r \sin (\omega t))\), as seen in the diagram below.

We have just argued the following.

Equation 0.1. Suppose an object is traveling in a circular path of radius \(r\) centered at the origin with constant angular velocity \(\omega \). If \(t=0\) corresponds to the point \((r,0)\), then the \(x\) and \(y\) coordinates of the object are functions of \(t\) and are given by \(x = r \cos (\omega t)\) and \(y = r \sin (\omega t)\). Here, \(\omega > 0\) indicates a counter-clockwise direction and \(\omega < 0\) indicates a clockwise direction.