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In section TheCircularFunctionsSineandCosine, we extended the notion of \(\sin (\theta )\) and \(\cos (\theta )\) from acute angles to any angles using the coordinate values of points on the Unit
Circle. In total, there are six circular functions, as listed below.
The Circular Functions: Suppose an angle \(\theta \) is graphed in standard position.
Let \(P(x,y)\) be the point of intersection of the terminal side of \(P\) and the Unit Circle.
The sine of \(\theta \), denoted \(\sin (\theta )\), is defined by \(\sin (\theta ) = y\).
The cosine of \(\theta \), denoted \(\cos (\theta )\), is defined by \(\cos (\theta ) = x\).
The tangent of \(\theta \), denoted \(\tan (\theta )\), is defined by \(\tan (\theta ) = \frac {y}{x}\), provided \(x \neq 0\).
The secant of \(\theta \), denoted \(\sec (\theta )\), is defined by \(\sec (\theta ) = \frac {1}{x}\), provided \(x \neq 0\).
The cosecant of \(\theta \), denoted \(\csc (\theta )\), is defined by \(\csc (\theta ) = \frac {1}{y}\), provided \(y \neq 0\).
The cotangent of \(\theta \), denoted \(\cot (\theta )\), is defined by \(\cot (\theta ) = \frac {x}{y}\), provided \(y \neq 0\).
While we left the history of the name ‘sine’ as an interesting research project in Section TheCircularFunctionsSineandCosine,we take a slight detour here to
explain the origin of the names ‘tangent’ and ‘secant.’
Consider the acute angle \(\theta \) in standard position sketched in the diagram below.
As usual, \(P(x,y)\) denotes the point on the terminal side of \(\theta \) which lies on the Unit Circle, but we also consider the point \(Q(1,y')\), the point on
the terminal side of \(\theta \) which lies on the vertical line \(x=1\).
The word ‘tangent’ comes from the Latin meaning ‘to touch,’ and for this reason, the line \(x=1\) is called a tangent line to the Unit
Circle since it intersects, or ‘touches’, the circle at only one point, namely \((1,0)\).
Dropping perpendiculars from \(P\) and \(Q\) creates a pair of similar triangles \(\Delta OPA\) and \(\Delta OQB\). Hence the corresponding sides are proportional.
We get \(\frac {y'}{y} = \frac {1}{x}\) which gives \(y' = \frac {y}{x} = \tan (\theta )\).
We have just shown that for acute angles \(\theta \), \(\tan (\theta )\) is the \(y\)-coordinate of the point on the terminal side of \(\theta \) which lies on the line \(x = 1\) which
is tangent to the Unit Circle.
The word ‘secant’ means ‘to cut’, so a secant line is any line that ‘cuts through’ a circle at two points. (Compare this with
the definition given in Section AverageRateofChange.) The line containing the terminal side of \(\theta \) (not just the terminal side itself) is one such secant
line since it intersects the Unit Circle in Quadrants I and III.
With the point \(P\) lying on the Unit Circle, the length of the hypotenuse of \(\Delta OPA\) is \(1\). If we let \(h\) denote the length of the hypotenuse of \(\Delta OQB\),
we have from similar triangles that \(\frac {h}{1} = \frac {1}{x}\), or \(h = \frac {1}{x} = \sec (\theta )\).
Hence for an acute angle \(\theta \), \(\sec (\theta )\) is the length of the line segment which lies on the secant line determined by the terminal side of \(\theta \)
and ‘cuts off’ the tangent line \(x=1\).
As we mentioned in Definition sinecosineunitcircledefn, the ‘co’ in ‘cosecant’ and ‘cotangent’ tie back to the concept of ‘co’mplementary angles and is
explained in detail in Section MoreTrigonometricIdentities.
Not only do these observations help explain the names of these functions, they serve as the basis for a fundamental inequality
needed for Calculus which we’ll explore in the Exercises.
Of the six circular functions, only sine and cosine are defined for all angles \(\theta \). Since \(x = \cos (\theta ) \) and \(y = \sin (\theta ) \) in Definition circularfunctions, it is customary to
rephrase the remaining four circular functions Definition circularfunctions in terms of sine and cosine.
We call the equations listed in Theorem recipquotididentities since they are relationships which are true regardless of the values of \(\theta \).
This is in contrast to conditional equations such as \(\sin (\theta ) = 1\) which are true for only some values of \(\theta \). We will study identities more
extensively in Sections FundamentalTrigonometricIdentities and MoreTrigonometricIdentities.
While the Reciprocal and Quotient Identities presented in Theorem recipquotid allow us to always reduce problems involving secant,
cosecant, tangent and cotangent to problems involving sine and cosine, it is not always convenient to do so. (As we shall see shortly, when solving equations involving secant and cosecant, we usually convert back to
cosines and sines. However, when solving for tangent or cotangent, we usually stick with what we’re dealt.) It
is worth taking the time to memorize the tangent and cotangent values of the common angles summarized
below.
Coupling Theorem recipquotid with the Reference Angle Theorem, Theorem refanglethm, we get the following.
Generalized Reference Angle Theorem. The values of the circular functions of an angle, if they exist, are the same, up to a
sign, of the corresponding circular functions of its reference angle.
More specifically, if \(\alpha \) is the reference angle for \(\theta \), then:
We have two ways to proceed to determine \(\tan (225^{\circ })\). First, we can use Theorem recipquotid and note that \(\tan (225^{\circ }) = \frac {\sin (225^{\circ })}{\cos (225^{\circ })}\). Since \(\sin (225^{\circ }) = \cos (225^{\circ }) = -\frac {\sqrt {2}}{2}\), \(\tan (225^{\circ }) = 1\).
Another way to proceed is to note that \(225^{\circ }\) has a reference angle of \(45^{\circ }\). Per Theorem genrefanglethm, \(\tan (225^{\circ }) = \pm \tan (45^{\circ }) = \pm 1\). Since \(225^{\circ }\) is a Quadrant III
angle, where both the \(x\) and \(y\) coordinates of points are both negative, and tangent is defined as the ratio
of coordinates \(\frac {y}{x}\), we know \(\tan (225^{\circ }) > 0\). Hence, \(\tan (225^{\circ }) = 1\).
As with the previous example, we have two ways to proceed. Using Theorem recipquotid, we have \(\cot \left (-\frac {7 \pi }{6} \right ) = \frac {\cos \left (-\frac {7 \pi }{6} \right )}{\sin \left (-\frac {7 \pi }{6} \right )}\). Since \(\cos \left (-\frac {7 \pi }{6} \right ) = -\frac {\sqrt {3}}{2}\) and \(\sin \left (-\frac {7 \pi }{6} \right ) = \frac {1}{2}\), we
get \(\cot \left (-\frac {7 \pi }{6} \right ) = - \sqrt {3}\).
Alternatively, we note \(-\frac {7 \pi }{6}\) is a Quadrant II angle with reference angle \(\frac {\pi }{6}\). Hence, Theorem genrefanglethm tells us \(\cot \left (-\frac {7 \pi }{6} \right ) = \pm \cot \left (\frac {\pi }{6} \right ) = \pm \sqrt {3}\). Since \(-\frac {7 \pi }{6}\) is a
Quadrant II angle, where the \(x\) and \(y\) coordinates have different signs, and cotangent is defined as the ratio
of coordinates \(\frac {x}{y}\), we know \(\cot \left (-\frac {7 \pi }{6} \right )<0\). Hence, \(\cot \left (-\frac {7 \pi }{6} \right ) = -\sqrt {3}\).
To solve \(\sec (\theta ) = 2\), we convert to cosines and get \(\frac {1}{\cos (\theta )} = 2\) or \(\cos (\theta ) = \frac {1}{2}\). This is the exact same equation we solved in Example solveforangle,
number cosineishalf, so we know the answer is: \(\theta = \frac {\pi }{3} + 2\pi k\) or \(\theta = \frac {5\pi }{3} + 2\pi k\) for integers \(k\).
From the table of common values, we see \(\tan \left (\frac {\pi }{3}\right ) = \sqrt {3}\). According to Theorem genrefanglethm, we know the solutions to \(\tan (\theta ) = \sqrt {3}\) must,
therefore, have a reference angle of \(\frac {\pi }{3}\).
To find the quadrants in which our solutions lie, we note that tangent is defined as the ratio \(\frac {y}{x}\) of points \((x,y)\)
on the Unit Circle. Hence, tangent is positive when \(x\) and \(y\) have the same sign (i.e., when they are both
positive or both negative.) This happens in Quadrants I and III.
In Quadrant I, we get the solutions: \(\theta = \frac {\pi }{3} + 2\pi k\) for integers \(k\), and for Quadrant III, we get \(\theta = \frac {4\pi }{3} + 2\pi k\) for integers \(k\). While these
descriptions of the solutions are correct, they can be combined into one list as \(\theta = \frac {\pi }{3} + \pi k\) for integers \(k\). The latter
form of the solution is best understood looking at the geometry of the situation in the diagram below. (See Example solveforangle number cosineiszero in Section TheCircularFunctionsSineandCosine for another example of this kind of simplification of the solution.)
From the table of common values, we see that \(\frac {\pi }{4}\) has a cotangent of \(1\), which means the solutions to \(\cot (\theta ) = -1\) have a
reference angle of \(\frac {\pi }{4}\).
To find the quadrants in which our solutions lie, we note that \(\cot (\theta ) = \frac {x}{y}\) for a point \((x,y)\) on the Unit Circle where \(y \neq 0\). If \(\cot (\theta )\) is negative,
then \(x\) and \(y\) must have different signs (i.e., one positive and one negative.) Hence, our solutions lie in Quadrants II
and IV.
Our Quadrant II solution is \(\theta = \frac {3\pi }{4} + 2\pi k\), and for Quadrant IV, we get \(\theta = \frac {7\pi }{4} + 2\pi k\) for integers \(k\). As in the previous problem, we can combine
these solutions as: \(\theta = \frac {3\pi }{4} + \pi k\) for integers \(k\).
A few remarks about Example circularfunctionsex are in order. First note that the signs (\(\pm \)) of secant and cosecant are the same as the signs of
cosine and sine, respectively.
On the other hand, since tangent and cotangent are defined in terms of the ratios of coordinates \(x\) and \(y\), tangent and cotangent
are positive in Quadrants I and III (where both \(x\) and \(y\) have the same sign) and negative in Quadrants II and IV (where \(x\) and \(y\)
have opposite signs.)
The diagram below summarizes which circular functions are positive in which quadrants.
Also note it is no coincidence that both of our solutions to the equations involving tangent and cotangent in Example circularfunctionsex could be
simplified to just one list of angles differing by multiples of \(\pi \).
Indeed, any two angles that are \(\pi \) units apart will not only have the same reference angle, but points on their terminal sides on
the Unit Circle will be reflections through the origin, as illustrated below.
It follows that the tangent and cotangent of such angles (if defined) will be the same, which means the period of these function
is (at most) \(\pi \).
Using an argument similar to the one we used to establish the period of sine and cosine in Section GraphsofSineandCosine, we note that if \(\tan (x+p) = \tan (x)\) for all real
numbers \(x\), then, in particular, \(\tan (p) = \tan (0+p) = \tan (0) = 0\). Hence, \(p\) is a multiple of \(\pi \), and the smallest multiple of \(\pi \) is \(\pi \) itself.
Hence, the period of tangent (and cotangent) is \(\pi \), and we will see the consequences of this both when solving equations in this
section and when graphing these functions in Section GraphsofOtherCircularFunctions.
Suppose \(Q(x,y)\) is the point on the terminal side of an angle \(\theta \) (plotted in standard position) which lies on the circle of radius \(r\), \(x^2+y^2 = r^2\).
Then:
Suppose the terminal side of \(\theta \), when plotted in standard position, contains the point \(Q(3,4)\). Find the values of the six
circular functions of \(\theta \).
Suppose \(\theta \) is a Quadrant IV angle with \(\cot (\theta ) = -4\). Find the values of the five remaining circular functions of \(\theta \).
Find \(\sin \left (\theta \right )\), where \(\sec (\theta ) = -\sqrt {5}\) and \(\theta \) is a Quadrant II angle.
Find \(\cos \left (\theta \right )\), where \(\tan (\theta ) = 3\) and \(\pi < \theta < \frac {3\pi }{2}\).
Since \(x = 3\) and \(y=4\), from \(x^2+y^2 = r^2\), \((3)^2+(4)^2 = r^2\) so \(r^2 = 25\), or \(r = 5\). Theorem circularfunctionscircle tells us \(\sin (\theta ) = \frac {4}{5}\), \(\cos (\theta ) = \frac {3}{5}\), \(\tan (\theta ) = \frac {4}{3}\), \(\sec (\theta ) = \frac {5}{3}\), \(\csc (\theta ) = \frac {5}{4}\), and \(\cot (\theta ) = \frac {3}{4}\).
In order to use Theorem circularfunctionscircle, we need to find a point \(Q(x,y)\) which lies on the terminal side of \(\theta \), when \(\theta \) is plotted in standard
position.
We have that \(\cot (\theta ) = -4 = \frac {x}{y}\). Since \(\theta \) is a Quadrant IV angle, we also know \(x>0\) and \(y< 0\). Rewriting \(-4 = \frac {4}{-1}\), we choose (We could have just as
easily chosen \(x=8\) and \(y=-2\) - just so long as \(x>0\), \(y<0\) and \(\frac {x}{y} = -4\).)\(x = 4\) and \(y = -1\) so that \(r = \sqrt {x^2+y^2} = \sqrt {(4)^2 + (-1)^2} = \sqrt {17}\).
To find \(\sin (\theta )\) using Theorem circularfunctionscircle, we need to determine the \(y\)-coordinate of a point \(Q(x,y)\) on the terminal side of \(\theta \), when \(\theta \) is plotted in
standard position, and the corresponding radius \(r\).
Since \(\sec (\theta ) = \frac {r}{x}\) and \(r > 0\), we rewrite \(\sec (\theta ) = \frac {r}{x} = -\sqrt {5} = \frac {\sqrt {5}}{-1}\) and take \(r = \sqrt {5}\) and \(x = -1\).
To find \(y\), we substitute \(x=-1\) and \(r=\sqrt {5}\) into \(x^2+y^2 = r^2\) to get \((-1)^2+y^2=(\sqrt {5})^2\). We find \(y^2 = 4\) or \(y = \pm 2\). Since \(\theta \) is a Quadrant II angle, we select \(y = 2\).
We are told \(\tan (\theta ) = 3\) and \(\pi < \theta < \frac {3\pi }{2}\), so we know \(\theta \) is a Quadrant III angle.
To find \(\cos (\theta )\) using Theorem circularfunctionscircle, we need to find the \(x\)-coordinate of a point \(Q(x,y)\) on the terminal side of \(\theta \), when \(\theta \) is plotted in standard
position, and the corresponding radius, \(r\).
Since \(\tan (\theta ) = \frac {y}{x}\) and \(\theta \) is a Quadrant III angle, we rewrite \(\tan (\theta ) = 3 = \frac {-3}{-1} = \frac {y}{x}\) and choose \(x = -1\) and \(y = -3\). From \(x^2+y^2 = r^2\), we get \(r = \sqrt {10}\).
As we did in Section cosinesinebeyond, we may consider \(\tan (t)\), \(\sec (t)\), \(\csc (t)\), and \(\cot (t)\) as functions real numbers by associating each real number \(t\) with
an angle \(\theta \) measuring \(t\) radians as discussed on page ?? and using Definition circularfunctions, or, more generally, Theorem
circularfunctionscircle.
Alternatively, we could define each of these four functions in terms of \(f(t) = \sin (t)\) and \(g(t) = \cos (t)\) as demonstrated in Theorem recipquotid. For example, we
could simply define\(\sec (t) = \frac {1}{\cos (t)}\) so long as \(\cos (t) \neq 0\).
Either way, we have the means to explore these functions in greater detail. Before doing so, we’ll need practice with these
additional four circular functions courtesy of the Exercises.
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