# The Trigonometric Functions by Stefan Waner and Steven R. Costenoble

## This Section: 2. The Six Trigonometric Functions

2. The Six Trigonometric Functions

The two basic trigonometric functions are: sine (which we have already studied), and cosine. By taking ratios and reciprocals of these functions, we obtain four other functions, called tangent, secant, cosecant, and cotangent.

Cosine

Let us go back to the bicycle introduced in the preceding section, and recall that the sine of $t, \sin t,$ was defined as the $y-$coordinate of a marker on the wheel. The cosine of $t,$ denoted by $\cos t,$ is defined in almost the same way, except that this time, we use the $x-$coordinates of the marker on the wheel. (See the figure.)

$\cos t$ is defined as the $x-$coordinate of the point $P$ shown.

First notice that the coordinates of the point $P$ in the above diagram are $(\cos t, \sin t),$ and that the distance from $P$ to the origin is $1$ unit. From the distance formula in Chapter 8 of Calculus Applied to the Real World or Chapter 15 of Finite Mathematics and Calculus Applied to the Real World we have:

Square of the distance from $P$ to $(0, 0) = 1$
$(\sin t)^2 + (\cos t)^2 = 1$

We often write this as

$\sin^2t + \cos^2t = 1,$

ant so we have found a relationship between the sine and cosine function.

 Fundamental Trigonometric Identity $\sin^2t + \cos^2t = 1$

Let us now turn attention to the graph of the cosine function. The graph, as you might expect, is almost identical to that of the sine function, except for a "phase shift" (see the figure).

This gives the following new pair of identities.

 Further Relationships Between Sine and Cosine The cosine curve is obtained from the sine curve by shifting it to the left a distance of $\pi/2.$ Conversely, we can obtain the sine curve from the cosine curve by shifting it $\pi/2$ units to the right. $\cos t = \sin(t + \pi/2)$ $\sin t = \cos(t - \pi/2)$ Alternative formulation We can also obtain the \cosine curve by first inverting the sine curve vertically (replace $t$ by $-t$) and then shifting to the right a distance of $\pi/2.$ This gives us two alternative formulas (which are easier to remember) $\cos t = \sin(\pi/2 - t)$ $\sin t = \cos(\pi/2 - t)$

Question
Since we can model the cosine function with a sine function, who needs the cosine function anyway?

Technically, that is correct; we don't need the cosine function and we can get by with the sine function by itself. On the other hand, it is convenient to have the cosine function around, since it starts at its highest point, rather than zero.

 Modeling with the Cosine Function (General Cosine Curve) Note that the basepoint is at the higher point of the curve. All the constants have the same meaning as for the general sine curve: $A$ is the amplitude (the height of each peak above the baseline) $C$ is the vertical offset (height of the baseline) $P$ is the period or wavelength (the length of each cycle) $ω$ is the angular frequency, given by $ω = 2\pi/P$ $α$ is the phase shift (the horizontal offset of the basepoint; where the curve reaches its maximum)

Example 1 Cash Flows into Stock Funds

The annual cash flow into stock funds (measured as a percentage of total assets) has fluctuated in cycles of approximately 40 years since 1955, when it was at a high point. The highs were roughly $+15%$ of total assets, while lows were roughly $-10%$ of total assets.*

(a) Model this cash flow with a cosine function of the time $t$ in years with $t = 0$ representing 1955.
(b) Convert the answer in part (a) to a sine function model.

* Source: Investment Company Institute/The New York Times, February 2, 1997. p. F8.

Solution

(a) Cosine modeling is similar to sine modeling: We are seeking a function of the form

$P(t) = A \cos[ω(t-α)] + C.$

Amplitude $A$ and Vertical Offset $C:$
Since the cash flow fluctuates between $- 10%$ and $+15%,$ we see that $A = 12.5,$ and $C = 2.5.$

Period $P$:
This is given as $P = 40.$

Angular Frequency $ω:$
This is given by the formula

$ω = 2\pi/P = 2\pi/40 = \pi/20 \approx 0.157.$

Phase Shift $α:$
The basepoint is at the high point of the curve, and we are told that cash flow was at its high at $t = 0.$ Therefore, the basepoint occurs at $t = 0,$ and so $α = 0.$

Putting this together gives

$P(t) = A \cos[ω(t-α)] + C$
$= 12.5\cos(0.157t) + 2.5,$

where t is time in years.

(b) To convert between a sine and cosine model, we can use the relations given above. Here, let us use the formula

$\cos x = \sin(x + \pi/2).$

Therefore,

$P(t) = 12.5\cos(0.157t) + 2.5$
$= 12.5\sin(0.157t + \pi/2) + 2.5.$

The Other Trigonometric Functions

As we said above, we can take ratios and reciprocals of sine and cosine to obtain four new functions. Here they are.

Tangent, Cotangent, Secant, and Cosecant
 $\tan x = \frac{\sin x}{\cos x}$ tangent $\cotan x = \frac{\cos x}{\sin x} = \frac{1}{\tan x}$ cotangent $\sec x = \frac{1}{\cos x}$ secant $\cosec x = \frac{1}{\sin x}$ cosecant

Example 2

Use technology to graph the curve $y = \sec x$ for $-2\pi ≤ x ≤ 2\pi$

Solution

Since

$\sec x = 1/\cos x,$

we can enter this function as

	$Y_1 = 1/\cos(x).$


To set the window, let us use $-2\pi ≤ x ≤ 2\pi,$ and $-7 ≤ y ≤ 7.$ Here is the graph we obtain.

Question
What are the vertical lines doing here?

Since we defined the secant function as $\sec x = 1/\cos x,$ we know that it is not defined when the denominator is zero. That is, when

$\cos x = 0.$

Consulting the graph of $\cos x,$ we find that this occurs when $x = ±\pi/2, ±3\pi/2, ±5\pi/2,...$

Therefore, these values are not in the domain of the secant function. Further, as $x$ approaches these values, $\sec x$ becomes very large numerically, but changes sign when we cross these values, causing the graphing calculator to make sudden jumps from large negative values of y to large positive values. Thus, the vertical lines are asymptotes.

If you have studied the section on limits in Chapter 3 of Calculus Applied to the Real World, or Chapter 10 of Finite Mathematics and Calculus Applied to the Real World, you will recognize this phenomenon in terms of limits; For instance,

$x$$\pi/2^{-} \sec x = ∞ x$$\pi/2^{+}$ $\sec x = -∞$

Before we go on...

Here are the graphs of all four of these functions. You might try to reproduce them and think about the asymptotes

$\tan x = \sin x/\cos x$

$\cotan x = \cos x/\sin x$

$\sec x = 1/\cos x$

$\cosec x = 1/\sin x$

The Trig Functions as Ratios in a Right Triangle

Let us go back to the figure that defines the sine and cosine, but this time, let us think of these two quantities as lengths of sides of a right triangle:

We are also thinking of the quantity t as a measure of the angle shown rather than the length of an arc. Looking at the figure, we find that

 $\sin t =$ length of side opposite the angle $t = \frac{\text{opposite}}{1} = \frac{\text{opposite}}{\text{\text{hypotenuse}}}$ $\cos t =$ length of side adjacent to the angle $t = \frac{\text{adjacent}}{1} = \frac{\text{adjacent}}{\text{\text{hypotenuse}}}$
 $\tan t = \frac{\sin t}{\cos t} = \frac{\text{opposite}}{\text{adjacent}}$

This gives us the following six formulas

The Trigonometric Functions as Ratios in a Right Triangle

 Defining Formula Ratio in Right Triangle $\sin t = y-$coordinate of point $P$ $\sin t = \frac{\text{opposite}}{\text{hypotenuse}}$ $\cos t = x-$coordinate of point $P$ $\cos t = \frac{\text{adjacent}}{\text{\text{hypotenuse}}}$ $\tan t = \frac{\sin t}{\cos t}$ $\tan t = \frac{\text{opposite}}{\text{adjacent}}$ $\cotan t = \frac{\cos t}{\sin t}$ $\cotan t = \frac{\text{adjacent}}{\text{opposite}}$ $\sec t = \frac{1}{\cos t}$ $\sec t = \frac{\text{\text{hypotenuse}}}{\text{adjacent}}$ $\cosec t = \frac{1}{\sin t}$ $\cosec t = \frac{\text{hypotenuse}}{\text{opposite}}$

We would welcome comments and suggestions for improving this resource.

Mail us at:
 Stefan Waner (matszw@hofstra.edu) Steven R. Costenoble (matsrc@hofstra.edu)

Last Updated: March, 1997