Trigonometric Functions

This section deals with trigonometric functions. These functions are some of the most important and applicable in mathematics.

Right Angle Trigonometry

We will start with the basic ideas of right angle trigonometry. Given a right triangle with angle x, as seen in the figure, we can define six ratios.

wpeF.gif (1282 bytes)

These six ratios are given the names sine, cosine, tangent, cotangent, secant, and cosecant and are defined as follows:

Note that all of these are functions of x, as x changes so does the ratio. These definitions immediately give rise to several relationships: . These relationships show that the functions we should be most concerned about are sine and cosine. Also, using the Pythagorean Theorem we come up with a simple but useful identity, sin2x+cos2x=1. (More identities) (The notation sin2x means the same thing as (sinx)2. The notation sinx2 means the same thing as sin(x2).) From these simple definitions we are able to solve some simple problems.

  1. The angle q is an acute angle of a right triangle. Solve the following problem by drawing an appropriate right triangle. Do not use a calculator.
  2. Find sinq and cosq given that tanq =3.

    Solution:

    The definition of tangent is so we know that the tangent must be =tanq . Drawing our triangle:

    wpe16.gif (1530 bytes)

    Using the Pythagorean theorem we can find the length of the hypotenuse, which turns out to be . Using the definition of sine we can find sinq . Sinq ==. Likewise we can do the same for cosine. The definition of cosq == .

    Circular Trigonometry

    The most useful view of the trigonometric functions is on the unit circle, the circle centered at the origin with radius 1. This relates to the above definitions, but gives us much more power in understanding the functions. We are going to do most of our work on the unit circle. We must first understand the concept of radian measure for angles. We know that the unit circle has circumference equal to 2p . We define and angle of 1 radian to be the angle measured at the center of the unit circle, beginning at the x-axis and moving counterclockwise, that cuts off an arc of length 1.

    wpe17.gif (1798 bytes)

    Thus 2 radians cut off an arc of length 2 and –1 radian cuts off an arc of length 1 in the clockwise direction. Thus one full time around the circle is 2p radians. Thus we have a relationship between radians and degrees. 2p radians = 360 degrees, or p radians=180 degrees. This gives us the following standard relationships:

    degrees

    radians

    0

    0

    30

    45

    60

    90

    120

    135

    150

    180

    p

    This table could be continued up to 2p radians, or in fact beyond 2p or in the negative direction. You should get used to radians, because radians will be our primary measure of angles.

    Considering the unit circle, drawn on the x-y plane, and the triangles given above we get an alternative, and often more useful, definition for sine and cosine. The cosine of the angle t is the x-coordinate on the circle at that angle. Similarly the sine of the angle t is the y-coordinate on the circle. See the following figure.

    wpe18.gif (2317 bytes)

    Using known ratios for sides in the 30-60-90 triangle and 45-45-90 triangle we can find the values for sine and cosine for all of these angles, as given in the following figure.

    Now, if you will take note of the symmetries of the above figure, you will see that you don’t need to memorize the whole figure. The first quadrant has all of the information you need. From this information you can reconstruct the rest just by knowing which of x and y are negative in the quadrant you are interested in. Thus we have values for a number of standard angles for the sine and cosine functions. From these it is easy to find values for the remaining trigonometric functions for these angles.

    Now let us consider the graphs of the trigonometric functions. First consider the sine function. If we begin at an angle of t=0 then sin0=0. As we increase t toward we find that sint is increasing toward 1. Note that it increases quickly at first and then more slowly until it reaches 1. As t continues to increase toward p we see that sint decreases back toward 0, at exactly the reverse rate as it increases. Now as t continues to increase we note that sint is now negative, but behaving in basically the same manner. Thus we can see the graph of f(t)=sint from 0 to 2p .

    From this point on we see that the y-coordinates are going to repeat the same pattern as we move around the circle. Hence we see that the sine function repeats itself endlessly.

    Thus, if we know one cycle of the function, we know the whole function. The time needed for the function to complete one cycle is called the period of the function. This applies to all trigonometric functions. Thus the period of the function f(x)=sinx is 2p . One other interesting value associated with trigonometric functions is the amplitude. The amplitude is half the distance between the maximum and minimum values (if they exist). Thus the amplitude of f(x)=sinx is 1, since its’ maximum is 1 and its’ minimum is –1. Now, it is easy to see that the behavior is the cosine function is similar to that of the sine function, except that it starts at a height of 1, rather than 0. The following graph contains both f(x)=sinx (black) and g(x)=cosx (blue).

    From this picture it is easy to see that the cosine is the same as the sine, just shifted units to the left. Thus we have the identity . It is also easy to see from this f(x)=cosx has period 2p and amplitude 1. We can also look at the graph of the tangent function.

    Several things are clear from this picture. First, f(x)=tanx is not defined at x=, , , and so has vertical asymptotes at these points. This makes sense since tanx= and cosx =0 at exactly these points. We can also see that f(x)=tanx has period p and the amplitude is not defined. The graphs of the reciprocal functions, cotangent, secant, and cosecant will be shown in blue, with their related functions, tangent, cosine, and sine shown in red. First we have tangent and cotangent.

    Next cosine and secant.

    And finally, sine and cosecant.

    We now have the graphs of the six basic trigonometric functions. We are now able to solve some simple equations.

  3. Find all values of q (in radians) that satisfy the given equation. Do not use a calculator.
  1. tanq = -1
  2. cosq =

Solution:

    1. To solve this all we have to do is find all the angles whose tangent is –1. In this case q = . But we know that the tangent function repeats or has a period of . Therefore if we add p to q we get , which is another answer. We could add any number of periods or even subtract periods and have the same answer. The way we write this is q =, n=0,1,2,… and this gives us all possible angles with a tangent of –1.
    2. We solve this part the same way we did part a. The basic angles q with cosq = are q =and q =. So far we have found only two solutions, but there are in fact infinitely many. But the period for cosine is 2p so we have q = and q =,
    3. n=0,1,2,…

We will now examine the more general forms of the trigonometric functions f(x)=Asin(Bx+C)+D and g(x)=Acos(Bx+C)+D. Let us examine the effect of each of the constants A, B, C, and D.

As seen in the General Functions review section, the constant D shifts the function vertically. Notice that this does not have any effect on the amplitude or the period. Shown here are the graphs of f(x)=sinx (black) and g(x)=sinx+2 (blue).

Also, the constant C shifts the function horizontally. This is called a phase shift. Again note that this constant does not have any effect on the amplitude or the period. Also recall that a if C is negative then the function shifts to the right, while if C is positive the function shifts to the left. Shown here are f(x)=cos(x) (black) and g(x)= (blue).

That takes care of the two constants that just shift the function. Now, consider the constant A. Both sine and cosine have a maximum value of 1 and a minimum value of –1. Multiplying by A will make the maximum value be A and the minimum value be –A, if A is positive. Otherwise –A is the maximum and A is the minimum. In either case the amplitude will become |A|. For example, let f(x)=cos(x) (black) and g(x)=3cos(x) (blue) and h(x)=-2cos(x) (red).

We can see that g(x) is 3 times as tall as f(x), and h(x) is twice as tall as f(x), but is flipped across the x-axis. Hence, f has amplitude 1, g has amplitude 3 and h has amplitude 2. So the absolute value of A gives us the amplitude.

Finally, B affects the period. If B is larger than 1, then the function will move through one complete cycle more quickly. Hence the period will decrease. If B is between 0 and 1 the function will complete the cycle more slowly and so the period will increase. In fact, the period of the function is given by . Let f(x)=sin(x) (black), g(x)=sin(3x) (blue), and h(x)= (red).

We see that g(x) cycles 3 times in the time it takes f(x) to cycle once, so the period of g(x) is that of f(x), or . Also note that it takes h(x) twice as long to complete its’ cycle as f(x). Hence the period of h(x) is 2 times that of f(x), or .

Now let us put all of these together. Let f(x)=2cos(4x+1)-2. The amplitude of f is 2, the period of f is . The graph will then be a sine curve, with amplitude 2, period , shifted 1 to the left and down 2.

Here are a couple more examples.

  1. Consider the function y=5+ cos(3x).
  1. What is its amplitude?
  2. What is its period?
  3. Sketch its graph.

Solution:

    1. The amplitude is 1.
    2. The period is.
    3. To sketch the graph we now have all the necessary information. We know that we are dealing with the cosine function, which goes from 1 to -1 but we are adding 5 so it now goes from 6 to 4, and we know the period is . So we start at (0,6) and we mark it. Then we can mark as the end point because that is one period. We know that half way between 0 and we will be through half a period, which puts us at

. Now we can draw the curve.

 

  1. Find a possible formula for the following graph.
  2. Solution:

    From looking at the graph we can deduce several things. First we can see that it starts at 0 which is just what the sine function does. The amplitude is 4 since that is half the height from the lowest point to the highest point. But because the curve goes in the opposite direction it must be a negative 4. We can see that the period for this function is p instead of the normal 2p, so we know that there is some multiplier on x in the function. Putting all this information together we come up with f(x)=-4 sin(2x).

    Inverse Trigonometric Functions

    On occasion, you may need to find a number with a given sine. For example, you might want ot find x such that sinx=0 or such that sinx=.7. The first of these equations can be solved by inspection, by noting that the sine of x is the y-coordinate on the unit circle. The only angles with y-coordinate 0 are x=0, p , 2p , , and hence these are the solution to this equation. The second equation is more difficult, but may be done with a calculator, but there are still infinitely many solutions. They may be found as x=.775397, 2.366195, .775397 2p , 2/366195 2p , . (Note that these angles are in radians.) If you try this with your calculator you will see that you get only one number, x=0 for the former and x=.775397 for the later. We can see that what we are looking for is an inverse of the sine function. To understand this we must first examine the sine function. Consider the following graph of f(x)=sinx.

    Note that the blue line intersects the graph in multiple places, in fact in infinitely many places. In the above example, the blue line is the line y=.7 so the places where the line intersects the graph of y=sinx are exactly the solutions to sinx=.7. You can see then why we get infinitely many solutions. Thus f(x)=sinx does not pass the horizontal line test and so does not have an inverse. In order to get an inverse we must restrict the domain of the original function in order to pass the horizontal line test. The traditional restriction for sine is , giving the following graph.

    Now we have a function that passes the horizontal line test and so we may define the inverse. The inverse sine, usually written "arcsin" or "sin-1" is defined as follows:

    For -1 y 1, arcsiny=x means sinx=y with . Note that sin-1x DOES NOT mean . If we wish to use exponents to express fractions we will use the notation (sinx)-1=. Since we now have an inverse function, we can easily find the graph of it, by reflecting it across the line y=x. Here we see f(x)=sinx (black) and g(x)=arcsinx (blue).

    Similarly, cosine must be restricted to 0 x p and tangent is restricted to . We see here the graph of the inverse tangent.

    From these we can find all solutions to the original equations. First we see that given any solution, another solution can be obtained by adding or subtracting the period of the function. Hence, in the above example with sine we add and subtract 2p . Also, we can see from the unit circle and any given value for an x-coordinate or a y-coordinate occurs twice on the circle. So we see, in our previous example, that we have one point located in the first quadrant and another in the second quadrant that both have y-coordinate of .7. By using the symmetry of the circle we can see that if we know the angle x giving the point in the first quadrant, then the angle for the point in the second quadrant must be p -x. Similarly, if the angle is in the fourth quadrant then the corresponding angle is in the third quadrant and is still p -x. Thus we have may find both angles, given the one from the calculator. This is how we find the two angles .775397 and 2.366195. In a similar fashion, if we are using the inverse cosine function we can use the angle x given by the calculator along with the angle –x, or 2p -x if you prefer, and then add and subtract multiples of 2p. We will end with a sample problem. You may also want to review example 2.

  3. a.) Using a calculator set in radians, make a table of values, to two decimal places, of f(x)=arcsin(x), for x= -1, -0.8, -0.6,…, 0,…, 0.8, 1. (Arcsine is denoted by sin-1 on most calculators).

b.) Sketch f(x)= arcsin(x). State the domain and range of f.

Solution:

  1. To do this all we need to do is put in the given values into a calculator and find the arcsin.
  2. X

    Arcsin(x)

    -1 -1.57
    -0.8 -0.93
    -0.6 -0.64
    -0.4 -0.41
    -0.2 -0.20
    0 0
    0.2 0.20
    0.4 0.41
    0.6 0.64
    0.8 0.93
    1 1.57

     

  3. Using the points from part a we can get a general picture of what arcsin(x) looks like.

The domain is (-1,1) and the range is .

Homework

Contents    General Functions    Inverses     Exponentials and Logarithms    Trigonometric Identities