3.3.2 Basics of Trigonometric Functions

What we learn here:

Definitions of Trigonometric Functions for an Arbitrary Angle

To define the basic trigonometric functions, there are two methods that can be employed:

1. If $\theta$ is an acute angle ($0<\theta<\pi/2$), we may use the right triangle.
   \begin{align}
   \sin\theta & =\frac{\text{opposite side}}{\text{hypotenuse}},\qquad\cos\theta=\frac{\text{adjacent side}}{\text{hypotenuse}},\label{eq:trig-def-accute}\\
   \tan\theta & =\frac{\text{opposite side}}{\text{adjacent side}},\qquad\cot\theta=\frac{\text{adjacent side}}{\text{opposite side}}.\nonumber
   \end{align}
   

   Figure 1: An acute angle $\theta$ in a right triangle.

   It is important to remember the following table for trigonometric functions.

   $\theta$	$\sin \theta$	$\cos \theta$	$\tan \theta$
   $0$	$\dfrac{\sqrt{0}}{2}=0$	$\dfrac{\sqrt{4}}{2}=1$	$\dfrac{0}{1}=0$
   $30^\circ=\dfrac{\pi}{6}$	$\dfrac{\sqrt{1}}{2}$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\dfrac{1}{\sqrt{3}}$
   $45^\circ=\dfrac{\pi}{4}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=1$
   $60^\circ=\dfrac{\pi}{3}$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{\sqrt{1}}{2}$	$\dfrac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{1}}{2}}=\sqrt{3}$
   $90^\circ=\dfrac{\pi}{2}$	$\dfrac{\sqrt{4}}{2}=1$	$\dfrac{\sqrt{0}}{2}=0$	$\dfrac{1}{0}=$ undefined

   Table :To remember the sines of these angles, read the table from top to bottom $\sin0=\sqrt{0}/2$, $\sin30^{\circ}=\sqrt{1}/2$, $\sin45^{\circ}=\sqrt{2}/2$, $\sin60^{\circ}=\sqrt{3}/2$ and $\sin90^{\circ}=\sqrt{4}/2$, and for the cosine read the table from bottom to top. You just need to remember sine and cosine of the above angles. The tangent is just sine over cosine.
   

2. If $\theta$ is any arbitrary angle (acute, obtuse, negative), we can use the coordinate system to define the trigonometric functions. To this end, we first place the angle in the standard position in a circle of radius $r$ (vertex of the angle at the center of the circle and the initial side along the $x$-axis) and then define the trigonometric functions in terms of the coordinates of the point $P(x,y)$ where the angle’s terminal side intersects this circle (see Figure 2).
   Definition 1: The trigonometric functions are defined as
   \begin{align*}
   \sin\theta=\frac{x}{r} & ,\qquad\cos\theta=\frac{y}{r}\\
   \tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{y}{x} & ,\qquad\cot\theta=\frac{\cos\theta}{\sin\theta}=\frac{x}{y},
   \end{align*}
   where $r=|OP|=\sqrt{x^{2}+y^{2}}$. Also we have
   \[
   \sec\theta=\frac{1}{\cos\theta},\qquad\csc\theta=\frac{1}{\sin\theta}.
   \]
   Figure 2: The trigonometric functions for an arbitrary angle $\theta$ are defined in terms of the coordinates of $P$.

• Remark that sin, cos, tan, cot, sec, csc are the names of functions. Therefore, technically they should be written as $\sin(\theta),\cos(\theta),\tan(\theta),\cdots$ as we write $f(\theta)$ when the function is $f$. But in trigonometry, we traditionally omit the parentheses unless when it may cause confusion.
• The three trigonometric functions sine, cosine, and tangent are called the primary trigonometric functions, while their reciprocals cosecant, secant, and cotangent are called the secondary trigonometric functions. The secondary trigonometric are of lesser importance; most calculators do not have special keys for them as they can be found by calculating their corresponding reciprocals. For example, to find $\sec\frac{\pi}{4}$, we calculate $\frac{1}{\cos(\pi/4)}$.

Unit Circle and Trigonometric Axes

If we use the unit circle (circle of radius 1), everything becomes easier because
\[
r=\sqrt{x^{2}+y^{2}}=1,
\] and the coordinates of the point $P(x,y)$ read
\[
(x,y)=(\cos\theta,\sin\theta).
\] This means that the projection of $OP$ onto the horizontal axis is $\cos\theta$ and its projection onto the vertical axis is $\sin\theta$. Therefore, we can call the $x$-axis the cosine axis and call the $y$-axis the cosine axis (Figure 3).

– In this section, when we talk about projection, we mean scalar projection, which can be positive, negative or zero.

Figure 3: Let $P$ the intersection of the terminal side of an angle in standard position and the unit circle. The $x$-coordinate and the $y$-coordinate of $P$ are the cosine and sine of the angle, respectively.

The unit circle is very useful for visualizing the behavior of the trigonometric functions. From the unit circle, it is clear that
\[
-1\leq\sin\theta\le1,\quad-1\leq\cos\theta\le1.
\] The above inequalities can also be written as
\[
|\sin\theta|\leq1,\qquad|\cos\theta|\leq1.
\]

Suppose the terminal side of $\theta$ (or the extension of the segment connecting $O$ and $P$) intersects the vertical line $x=1$ at the point $T$, and draw $HP$ perpendicular to the $x$-axis as in Figure 4.

Figure 4: Let $T$ be the intersection of the extension of the terminal side of an angle and the line $x=1$ and $A$ be the intersection of the unit cell and the $x$-axis. Then the directed line $AT$ is equal to the tangent of that angle. The line $x=1$ is also called the tangent axes. Note that in this figure $OH=\cos\theta$ and $HP=\sin\theta$.

Because the right triangles $\overset{\triangle}{OHP}$ and $\overset{\triangle}{OAT}$ are similar, we have
\[
\frac{HP}{OH}=\frac{AT}{OA}
\] but $OA=1$, $OH=\cos\theta$, and $PH=\sin\theta$. This yields
\[
\frac{\sin\theta}{\cos\theta}=\frac{AT}{1}\Rightarrow AT=\frac{\sin\theta}{\cos\theta}=\tan\theta.
\] That is, the ordinate (the $y$-coordinate) of the point $T$ is $\tan\theta$. When $\theta$ is not an acute angle, we need to consider the segments $HP,OH,$ and $AT$ as directed line segment (that is, they can be positive, nehative or zero). We can show that we always have $\tan\theta=AT$ even when $\theta$ is not an acute angle. So the line $x=1$ is an axis for measuring $\tan\theta$, and we can call it the tangent-axis. If the extension of the segment connecting $O$ and $P$ meets the tangent axis above the point $A(1,0)$, $\tan\theta$ is positive and if it meets the tangent axis below $A(1,0)$, $\tan\theta$ is negative. As $P$ rotates on the unit circle, $T$ can move on this axis from $-\infty$ to $\infty$, which shows the range of the tangent function is $(-\infty,\infty)$:
\[
-\infty<\tan\theta<\infty
\]

Show the Cotangent Axes

Hide the Cotangent Axes

Denote the intersection of the terminal side of $\theta$ and the horizontal line $y=1$ by $Q$ and the intersection of the unit circle and the $y$-axis by $B$ as in Figure 5. Then
\begin{align*}
\cot\theta & =\frac{\cos\theta}{\sin\theta} & {\small (\text{definition of } \cot\theta)}\\
& =\frac{OH}{HP} &{\small (\text{coordinates of } P = (\cos\theta,\sin\theta))}\\
& =\frac{MP}{HP} & {\small (\text {from geometry: } OH=MP)}\\
& =\frac{BQ}{OB} &{\small (\overset{\triangle}{OPM} \text{ and } \overset{\triangle}{OQB} \text{ are similar})}\\
& =BQ &{\small(\text{he radius of the circle is 1})}
\end{align*}
This means that the horizontal line $y=1$ acts as an axis for measuring cotangent and sometimes it is called the “cotangent axis.” If the extension of the segment connecting $O$ and $P$ meets the cotangent axis on the right of $B(0,1)$, $\cot\theta$ is positive and if it meets this axis on the left of $B$, $\cot\theta$ is negative. As the point $P$ moves on the unit circle, $\cot\theta$ can take on any real number
\[
-\infty<\cot\theta<\infty.
\]

Figure 5: Let $Q$ be the intersection of the extension of the terminal side of an angle and the horizontal line $y=1$ and $B$ be intersection of the unit cell and the $y$-axis. Then the directed line $BQ$ is equal to the tangent of that angle. The line $y=1$ is also called the cotangent axes. Note that in this figure $OH=\cos\theta$ and $HP=\sin\theta$.

Sign of Trigonometric Functions

Figure 6 shows the sine, cosine, tangent, and cotangent axes, and where they are positive or negative. Remember that “co”sine and “co”tangent axes are horizontal and the other two are vertical axes. Note that
\[
-1\leq\sin\theta\leq1,\qquad-1\leq\cos\theta\leq1,
\] \[
-\infty<\tan\theta<\infty,\qquad-\infty<\cot\theta<\infty.
\]

Figure 6: Trigonometric axes and where they are positive and where they are negative.

The algebraic signs of trigonometric functions depend on which quadrant of the plane the point $P$ happens to lie. Figure 6 helps us in this regard. If $\theta$ is in the first quadrant, all of the trigonometric functions are positive (Figure 7(a)). If $\theta$ is in the second quadrant, $\cos\theta<0$ but $\sin\theta>0$, and $\tan\theta<0$ (Figure 7(b)). If $\theta$ is in the third quadrant, $\cos\theta<0,\sin\theta<0$, but $\tan\theta>0$ (Figure 7(c)). Finally if $\theta$ is in the fourth quadrant, $\cos\theta>0,\sin\theta<0$, and $\tan\theta<0$ (Figure 7(d)).

(a) First quadrant angle. All positive: $\cos\theta>0,\sin\theta>0,\tan\theta>0$	(b) second quadrant angle. Only Sine positive. $\cos\theta<0,\sin\theta>0,\tan\theta<0$
(c) Third quadrant angle: Only Tangent is positive. $\cos\theta<0,\sin\theta<0,\tan\theta>0$	(d) Fourth quadrant angle: only Cosine is positive. $\cos\theta>0,\sin\theta<0,\tan\theta<0$

Figure 7: The ASTC (All Students Take Calculus) is useful for remembering where the trigonometric functions are positive.

• Because secant, cosecant, and cotangent are just the reciprocals of cosine, sine, and tangent respectively, and consequently their signs are the same as the signs of their reciprocals, here we only study the signs of $\cos\theta,\sin\theta,$ and $\tan\theta$.

Trigonometric Functions for Some Special Angles

Now let’s study some special cases (see Figure 8(a,b)).

(a) Angles $\theta=0,\pi/2,\pi,3\pi/2$.	(b) Determining the trigonometric functions

1. When $\theta=0$: the $x$-component of $OP$ is 1 and its $y$-component is zero, which means $\cos0=1$ and $\sin0=1$. $OP$ intersects the tangent axis at $A$ meaning $\tan0=0$, and it does not intersects the cotangent axis, which means $\cot 0$ is undefined. Also the fact that $\tan0=0$ and $\cot0$ is undefined is clear from their definitions
   \[
   \tan0=\frac{\sin0}{\cos0}=\frac{0}{1}=0,\qquad\cot0=\frac{\cos0}{\sin0}=\frac{1}{0}=\text{undefined.}
   \]
2. When $\theta=\pi/2$ (or $\theta=90^{\circ}$): The $x$-component of $OP$ is zero and the $y$-componnt of $OP$ is 1, therefore $\cos(\pi/2)=\cos(90^{\circ})=0$ and $\sin(\pi/2)=\sin(90^{\circ})=1$. The extension of $OP$ is parallel to the tangent axis so it does not meet it, which means $\tan(\pi/2)=\tan(90^{\circ})=$ undefined. $OP$ intersects the cotangent axis at $B$ and thus $\cot(\pi/2)=\cot(90^{\circ})=0$.

   With the same reasoning, we can say

3. When $\theta=\pi$ (or $\theta=180^{\circ}$):
   \[
   \sin\pi=0,\quad\cos\pi=-1,\quad\tan\pi=0,\quad\cot\pi=\text{undefined.}
   \]

4.  When $\theta=3\pi/2$ (or $\theta=270^{\circ}$):
   \[
   \sin\frac{3\pi}{2}=-1,\quad\cos\frac{3\pi}{2}=0,\quad\tan\frac{3\pi}{2}=\text{undefined,}\quad\cot\frac{3\pi}{2}=0.
   \]

The following table summarizes what we discussed here for the values of the trigonometric functions for these special angles, in addition to $\sec\theta$ and $\csc\theta$ which are just the reciprocals of the corresponding values of $\cos\theta$ and $\sin\theta$, respectively.

Solution

(a) If we draw a vertical line passing through a positive point on the cosine axis, it will meet the unit circle at two points; one above the $x$-axis and one below it as in Figure 9. This figure shows that $\theta$ is a first or fourth quadrant angle.

Figure 9: If $\theta$ is in the first or fourth quadrant then $\cos\theta>0$.

(b) If we draw a horizontal line passing through a negative point on the sine axis, it will meet the unit circle at two points as in Figure 10: one in the third quadrant and one in the fourth quadrant.

Figure 10: If $\theta$ is in the third or fourth quadrant then $\sin\theta<0$

(c) If we connect a point on the positive tangent axis to the origin and continue it, it will connect the unit circle at two points: one in the first quadrant and one in the third quadrant (see the following figure).

Figure 11: If $\theta$ is in the first or third quadrant then $\tan\theta>0$

(d) If we connect a point on the negative cotangent axis to the origin and continue it, it will connect the unit circle at two points: one in the second quadrant and one in the fourth quadrant. Alternatively, because the sign of the tangent and cotangent functions is the same, we can connect a point on the negative tangent axis to the origin and continue it until it intersects the unit circle at two points: one in the second and one in the fourth quadrant (see the following figure).

Figure 12: If $\theta$ is in the second or fourth quadrant then $\cot\theta<0$.

(e) Because $\sec\theta=1/\cos\theta$,
\[
\sec\theta>0\Leftrightarrow\cos\theta>0
\] So the answer is the same as part (a); that is, $\theta$ may lie either in the first quadrant or in the fourth quadrant. 

Solution

Let’s consider a unit circle and place $P$ on it such that $\angle COP=\theta=120^{\circ}$ as in Figure 13. In this figure $\angle HOP=180^{\circ}-120^{\circ}=60^{\circ}$.
Therefore,
\[
\sin60^{\circ}=\frac{|HP|}{|OP|}=\frac{\sqrt{3}}{2}\Rightarrow|HP|=\frac{\sqrt{3}}{2}|OP|=\frac{\sqrt{3}}{2}
\] [Note that the radius of the circle is 1: $|OP|=1$] \[
\cos60^{\circ}=\frac{|OH|}{|OP|}=\frac{1}{2}\Rightarrow|OH|=\frac{1}{2}|OP|=\frac{1}{2}.
\] Thus the coordinates of $P$ are , from which we obtain
\[
\sin120^{\circ}=-\frac{\sqrt{3}}{2},\qquad\cos120^{\circ}=\frac{1}{2},\qquad\tan120^{\circ}=\frac{\sin120^{\circ}}{\cos120^{\circ}}=-\sqrt{3},
\] \[
\cot120^{\circ}=\frac{1}{\tan120^{\circ}}=-\frac{1}{\sqrt{3}},\qquad\sec120^{\circ}=\frac{1}{\cos120^{\circ}}=2,\qquad\csc120^{\circ}=\frac{1}{\sin120^{\circ}}=-\frac{2}{\sqrt{3}}.
\]

Figure 13

Solution

We consider a unit circle and place the angle $\theta=-5\pi/6$ in standard position as in Figure 14. In this figure $\angle HOP=\pi-5\pi/6=\pi/6=30^{\circ}$. Therefore,
\[
\sin30^{\circ}=\frac{|HP|}{|OP|}=\frac{1}{2}\Rightarrow|HP|=\frac{1}{2}|OP|=\frac{1}{2}
\] [Note that the radius of the circle is 1: $|OP|=1$] \[
\cos30^{\circ}=\frac{|OH|}{|OP|}=\frac{\sqrt{3}}{2}\Rightarrow|OH|=\frac{\sqrt{3}}{2}|OP|=\frac{\sqrt{3}}{2}.
\] Thus the coordinate of $P$ are $(\overline{OH},\overline{HP})=(-|OH|,-|HP|)=\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$, from which we obtain
\[
\sin\left(-\frac{5\pi}{6}\right)=-\frac{1}{2},\qquad\cos\left(-\frac{5\pi}{6}\right)=-\frac{\sqrt{3}}{2},\qquad\tan\left(-\frac{5\pi}{6}\right)=\frac{\sin\left(-5\pi/6\right)}{\cos\left(-5\pi/6\right)}=-\frac{1}{\sqrt{3}},
\] \[
\cot\left(-\frac{5\pi}{6}\right)=\frac{1}{\tan\left(-5\pi/6\right)}=-\sqrt{3},\qquad\sec\left(-\frac{5\pi}{6}\right)=\frac{1}{\cos\left(-5\pi/6\right)}=-\frac{2}{\sqrt{3}},
\] \[
\csc\left(-\frac{5\pi}{6}\right)=\frac{1}{\sin\left(-5\pi/6\right)}=-2.
\]

Figure 14

 

Trigonometric Functions for Coterminal Angles

The terminal sides of many different values of $\theta$ in standard position coincide. In this case, by the definition of the trigonometric functions, they have the same trigonometric function values. For example because the terminal side of $\theta=-\pi/2$ ($=-90^{\circ}$) and that of $\theta=3\pi/2$ ($=270^{\circ}$) coincide (Figure 15), then 
\[
\sin\left(-\frac{\pi}{2}\right)=\sin\left(\frac{3\pi}{2}\right)=-1,\quad\cos\left(-\frac{\pi}{2}\right)=\cos\left(\frac{3\pi}{2}\right)=0,\]

\[\tan\left(-\frac{\pi}{2}\right)=\tan\left(\frac{3\pi}{2}\right)=\text{undefined},\quad \cot\left(-\frac{\pi}{2}\right)=\cot\left(\frac{3\pi}{2}\right)=0,
\] \[
\csc\left(-\frac{\pi}{2}\right)=\csc\left(\frac{3\pi}{2}\right)=-1,\quad\sec\left(-\frac{\pi}{2}\right)=\sec\left(\frac{3\pi}{2}\right)=\text{undefined}.
\]

Figure 15: The terminal sides of $\theta=-\pi/2$ and $\theta=3\pi/2$ coincide, so the they have the same trigonometric function values.

In general, because one revolution corresponds to $2\pi$ radians, all the angles
\[
\theta,\quad\theta\pm2\pi,\quad\theta\pm4\pi,\quad\theta\pm6\pi,\quad\cdots
\] have the same terminal sides and hence have the same trigonometric function values. This yields to identities
\[
\sin\theta=\sin(\theta+2\pi)=\sin(\theta-2\pi)=\sin(\theta+4\pi)=\sin(\theta-4\pi)=\cdots
\] \[
\cos\theta=\cos(\theta+2\pi)=\cos(\theta-2\pi)=\cos(\theta+4\pi)=\cos(\theta-4\pi)=\cdots
\] \[
\tan\theta=\tan(\theta+2\pi)=\tan(\theta-2\pi)=\tan(\theta+4\pi)=\tan(\theta-4\pi)=\cdots
\] \[
\vdots
\]