Recently in Honors Precalc, we have learned how to transform graphs of sine and cosine. Here, I will teach you how.

(Within this post, the sign for theta keeps changing to a question mark after saving or previewing so if you see a ? within an equation, it is the theta symbol. Sorry for the inconvenience.)

We will start by going over the factors that add to these transformations. These entitle vertical shift, amplitude, period, and phase displacement. On a graph, the vertical shift determines the location of the sinusoidal axis. The amplitude determines the distance the sine or cosine curve will travel from the sinusoidal axis. The period is what determines how long in the x direction the graph will be relative to the normal 360° for the parent function of sine and cosine. Finally, phase displacement identifies at which point the sine or cosine graph will begin. For example, since the cosine graph starts high, you will need to find a high point on the graph to know where the cosine graph begins. In terms of equations, here is where these factors fit in… 

a + b sin c (?+ d)              and              a + b cos c (? + d)

a: vertical shift, b: amplitude, c: period, and d: phase displacement

As for A and B, the numbers seen within the equation are exactly what are used to find those aspects of the graph. For C and D, there are changes. As for C, or the period, the number seen in the equation is 360/period. This is because the normal sin and cos graphs have a period of 360°. For D, or phase displacement, the place where the graph begins is simply the number theta must be for the parentheses to equal 0. With this information, you could truly say that in the equations, a + b sin c (? + d) and a + b cos c (? + d):

a = vertical shift

b = amplitude

360/c = period

phase displacement + d = 0

which simplifies down to:

-d = phase displacement

Now that you know the steps, it is time to try some examples.

#1: Find the graph corresponding with this equation: 6 + 10 cos 2 (?+90)

First things first, you should find the factors that effect the graph.

In this equation,vertical shift = 6, and amplitude = 10. To find period and phase displacement, you must use the equations from earlier. First, to find period, you must use 360/c = period. C in this equation is 2 which means that the period, 360/2, is 180°. This means that one cosine curve will take up 180°. Now, in order to find phase displacement, we use the equation phase displacement + d = 0 or -d = phase displacement. Since positive 90 is in place of d, we know the phase displacement must be -90 thanks to those equations.

With this information, we can now create a graph illustrating this equation. First, we can put our sinusoidal graph at 6 since that is our vertical shift. This is shown in the picture at the left. Next, we know that our graph begins at -90 since this is our phase change. Since our graph is of cosine, we know that it starts high. We can count from the sinusoidal axis up as many as our amplitutde. This will give us a first point of (-90, 16). Now we have to find our next point. This can be done by dividing our period by 4 since every high, middle, and low point is located at one of the 1/4 points. Since our period is 180°, we know that each point is 45° away from one another since 180/4 is 45. Since the second point of  cosine graph is always on the sinusoidal axis. This means that this point is 45 over from the x value of the first point and 10 down from the y value of the first point. So if the first point was (-90,16), the second point on the cosine graph will be (-45, 6). Then the next point is another 45 over on the x axis and at its low point for y which is 10 down from the sinusoidal axis. This makes the third point of this graph (0, -4). Now we must find the next point on the graph which in cosine is at the sinusoidal axis once again. This point is over 45 and since it was at its low point last point, it must go up 10. This point is (45, 6). Now for the last point of this cosine curve, you again move 45 over on the x axis and up 10 since cosine ends at its high point. This last point is (90, 16). Now you can graph it all and end up with the graph for 6 + 10 cos 2 (? + 90). Below you can see this graph.

#2: Find the equation from the graph below:

Now, we are going to want to do the same thing as the last problem but backwards. Let’s look at the equation a + b sin/cos c (? + d). We must find all of the values for a, b, c, d, and which trig function it is to create the proper equation. To find the vertical shift we need to find the y location of the sinusoidal axis. In this graph, the axis is at 13. This means a = 13. Next we can find the amplitude from the distances up and down from the sinusoidal axis. Since the axis is at 13 and the high point is at 28, we can find the amplitude by subtracting 13 from 28. This will give us an amplitude of 15. This means b = 15. Now, we must find the period. Since there is exactly 1 curve between 45 and 765, we can subtract 45 from 765 to find how long the curve is. This will give us a period of 720. Now, to find c, we must divide 360 by the period. 360/720 is 1/2. This means that c = 1/2. Now we must determine which trig function is being displayed in order to find the phase displacement. Since cosine starts high and sine starts at the sinusoidal axis, we can choose which one is illustrated in the graph. The first y value on this graph is 13 which is on the sinusoidal axis which makes it a sine graph. Now that we  know where the graph begins, we can find the phase displacement. Since the x vlaue of the first point is at 45, we know that 45 was our starting point. At this point ? is 45 so in order to make 0 inside of the parentheses, you must subtract 45. This means that d = -45. Overall, with this sine graph having a = 13, b = 15, c = 1/2, and d = -45, you can write out the equation: 13 + 15 sin 1/2 (? – 45)

So you know how to transform sine and cosine graphs. I hope I helped. Bye.

-Amy D!

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Sine and cosine are in accordance with the unit circle, a circle whose radius is equal to one. Thus the lengths of sine and cosine can not be greater than one.

representation of sine and cosine in quadrant 1 of the unit circle

The origin of the circle is represented at (0,0) on a coordinate graph. because the radius (the length from the center of the circle to any edge of the circle) reaches a maximum length of 1, we know that cosine’s maximum length is 1 as well. Likewise, the coordinate graph has negative lengths. But because the radius of the unit circle is equal to one, we know that any negative value will no exceed -1. so cosine is represented on the x-axis making sine represented on a the y-axis. Same rules apply in terms of positive and negative values.

the unit circle

To understand the values of sine and cosine you must recognize when both are positive and negative values.

Sine is positive in quadrants 1 and 2 because y values are positive in 1 and 2.

Sine is negative in quadrants 3 and 4 because y values are negative in those quadrants.

Cosine is positive in quadrants 1 and 4, and negative in quadrants 2 and 3 because the x values coincide with such quadrants.

Perhaps the most important values you must know for sine and cosine are given in the degree measures of 0, 30, 45, 60, and 90.

When:

*sqr = square root*

degree measure:          sine value:          cosine value:

            0                                     0                                  1

            30                                 1/2                        sqr(3)/2        

           45                             sqr(2)/2                   sqr(2)/2

          60                             sqr(3)/2                          1/2

          90                                    1                                    0

Knowing these values will allow you to find the length of sine or cosine in any quadrant. in order to do so, if not found in quadrant 1, you need a reference angle. you do this by taking the angle measure and subtracting it from the nearest horizontal angle value (either 180 or 360 degrees). It helps if you can recognize which degree measure is in which quadrant.

Ex.

sin(135) = ?

ask: is 135 closer to 180 or 360

       180

180 – 135 = 45.    Thus, the reference angle is 45.

sin(45) = sqr(2)/2

ask: which quadrant is 135 located in?

         quadrant 2

ask: is sine positive or negative in quadrant 2?

         positive

so the reference angle is 45

we know the sin(45) = sqr(2)/2

135 degrees is located in quadrant 2

sine is a positive value in quadrant 2

thus, the sin(135) = sqr(2)/2

by knowing the positive or negative lengths of sine or cosine in a given quadrant and referring to their values within quadrant one at degree measures 0, 30, 45, 60, and 90 you will be able to understand and comprehend the values of sine and cosine throughout the unit circle.

In many math classes, a c?ircle is defined to have 360°.  However, in more advanced classes, such as calculus, one uses radians instead of degrees.  A radian is, by definition, the number of radii that would fit around a circle’s arc for a given angle.  Since the radius of a circle and its circumference are directly related and we are using a circle’s radius to measure and angle, an angle measure of a smaller circle would have the same radian measure as one that is larger.

The circumference of a circle is equal to ??2?r, r being the radius.  Therefore, we can say that 360° = 2? radians.  This then means that 180° = ? radians.

From this information, we can easily find the radian measures of many common degree measures easily.  Since 90° is 1/2 of 180°, the radian measure of 90° must then be 1/2 the radian measuer of 180°, or ?/2.

45° is 1/2 of 90°, so the radian equivalent of 45° is ?/4.  30° is 1/6 of 180°, so the radian equivalent of 30° is ?/6.  Here is a list of other angles and their exact radian equivalent:

An easy way to convert from degrees to radians is to set up a proportion.  I personally like to use the following, although you could substitute any other pair of degree and radian measures that are equivilant:

Plug in whitchever measurement you have – degree or radian – and cross-multiply to solve for the missing measure.

One of the handy things about radian measure is that you can easily find a specific arc length.

For isntance, let’s say that ? in this circle is equal to 120°.  If we convert it to radians, we find that it is equal to 2?/3 radians.  This means that there are exactly 2?/3 radians in this arc length s. Therefore, we simply need to multiply the radian measure by the length of the radius to find the arc length.  In this case, let’s say that the radius is 5 units long.  Therefore, the arc length is 5·(2?/3), or 10?/3.