# Graphing Sine & Cosine

## Trigonometry - Graphing Sine & Cosine - How it Works - Video

### Models of Sine & Cosine Graphs Here we can use a to find the amplitude or the distance from one extrema (minimum or maximum) to the middle of the graph. We can use b to help use find the period, 2π / |b|. The period is one cycle. It doesn't matter where you start, but most people start at one extrema and go to the next extrema that is horizontal to the one that you first chose.

Also, a helps us to determine if we have a vertical shrink or vertical stretch. If |a| > 1, then we have a vertical stretch. If |a| < 1, then we have a vertical shrink.

And, b helps us to determine if we have horizontal shrink or horizontal stretch. If |b| > 1, then we have a horizontal shrink. If |b| < 1, then we have a horizontal stretch. Here we can use c to find the phase shift or the horizontal shift which tells us whether the new graph shifts right or left. If we grab a ruler and lay it parallel to the x-axis, any two corresponding points on the two graphs, will be the same distance horizontally. Lucky for us the when we shift right or left, the amplitude and the period will stay the same. Here we can use d to find the vertical shift which tells us whether the new graph shifts up or down. If we grab a ruler and lay it parallel to the y-axis, any two corresponding points on the two graphs, will be the same distance vertically. Lucky for us the when we shift up or down, the amplitude and the period will stay the same.

### Theorems ### Example 1 Example 1:

Here we labeled out parts.

a = 2 b = 3 c = 0 d = 0

Remember a helps us find the amplitude => |a| = |2| = 2 and we have a vertical stretch since |a| > 1.

Remember b helps us find the period => 2π / |b| = 2π / |3| = 2π / 3 and we have a horizontal shrink since |b|> 1 .

Remember c helps us find the phase shift, since c = 0, we don't have a phase shift.

Remember d helps us find the vertical shift, since d = 0, we don't have a vertical shift. Here we use inequality, 0 ≤ bx + c ≤ 2π , to find one cycle.

a = 2 b = 3 c = 0 d = 0

0 ≤ bx + c ≤ 2π

0 ≤ 3x + 0 ≤ 2π

0 ≤ 3x ≤ 2π

0 ≤ x ≤ 2π /3

Substituted for b and c.

Subtracted 0.

Divided each section by 3.

Now we have the starting point of one cycle 0 and the end point 2π /3.

Since we are dealing with sine, the starting point starts on the x-axis or the origin in this case.

1 - Draw a line until 2, the amplitude (make sure the maximum is in the middle of the starting point and where it crosses the x-axis)

2 - Start to go down and crosses the x-axis in the middle of the cycle

3 - Continue until -2, the amplitude, remember the amplitude is a distance so the |-2| is 2 (make sure the minimum is in the middle of where it crosses the x-axis and the end point)

4 - Continue until the end point

We have completed one cycle. Make sure to add more cycles to get a better feel on how it looks. Let's check to see if we did everything correctly by graphing y = sin(x), y = 2 * sin(x), and y = 2 * sin(3x).

The black line represents y = sin(x); the red line represents y = 2 * sin(x); the blue line represents y = 2 * sin(3x).

The red graph is taller and crosses the x-axis at the same spots as the black graph. That is because we have a vertical stretch since |a| > 1. The input value is not affected by the number 2 since sin(x) goes first. Since the 2 doubles everything, our graph is taller, which is why we have a vertical stretch.

Now let's look at the red and blue graph.

The blue graph is the same height as the red graph, that is because the amplitude is the same, but it does not cross the x-axis at the same spots because the period is not the same. Since we have the same amount of ups and downs in each graph, we can tell the period of the blue graph is smaller. Our b value affects how fast we go right and left, since |b| > 1 we have a horizontal shrink, which is why our period is smaller.

That is how we graph y = 2 * sin(3x).

### Example 2 Example 2:

Here we labeled out parts.

a = 3 b = 2 c = 7/5 d = 1

Remember a helps us find the amplitude => |a| = |3| = 3 and we have a vertical stretch since |a| > 1.

Remember b helps us find the period => 2π / |b| = 2π / |2| = π and we have a horizontal shrink since |b| > 1.

Remember c helps us find the phase shift => -c / b = - 7/5 / 2 = -7/10 so we have a phase shift left.

Remember d helps us find the vertical shift = 1. Here we use inequality, 0 ≤ bx + c ≤ 2π , to find one cycle.

a = 2 b = 3 c = 0 d = 0

0 ≤ bx + c ≤ 2π

0 ≤ 2x + 7/5 ≤ 2π

-7/52x ≤ 2π - 7/5

-7/10 ≤ x ≤ π - 7/10

Substituted for b and c.

Subtracted -7/5 to each section.

Divided each section by 2.

Now we have the starting point of one cycle -7/10 and the end point π - 7/10.

Since we are dealing with cosine, the starting point starts on the y-axis.

1 - We start at -7/10 and move our dot up 3, the amplitude

2 - Now we draw a line down and cross at the x-axis make sure it is in the middle of the starting point the middle of the cycle

3 - Continue until -3, the amplitude, remember the amplitude is a distance so the |-3| is 3 (make sure the minimum is in the middle of the cycle)

4 - Now we draw a line up and cross at the x-axis make sure it is in the middle of the cycle of the end point

5 - Continue until 3, the amplitude

We have completed one cycle. Now we have to worry about the letter d. Since d = 1 our next step is to move the graph up 1 since it is positive. So our first cycle is the green graph.

Once you get a feel, you can combine these last two steps. Make sure to add more cycles to get a better feel on how it looks. Let's check to see if we did everything correctly by graphing y = cos(θ), y = 3 * cos(θ), and y = 3 * cos().

The black line represents y = cos(θ); the red line represents y = 2 * cos(θ); the blue line represents y = 3 * cos().

The red graph is taller and crosses the x-axis at the same spots as the black graph. That is because we have a vertical stretch since |a| > 1. The input value is not affected by the number 3 since cos(x) goes first. Since the 3 triples everything, our graph is taller, which is why we have a vertical stretch.

Now let's look at the red and blue graph.

The blue graph is the same height as the red graph, that is because the amplitude is the same, but it does not cross the x-axis at the same spots because the period is not the same. Since we have the same amount of ups and downs in each graph, we can tell the period of the blue graph is smaller. Our b value affects how fast we go right and left, since |b| > 1 we have a horizontal shrink, which is why our period is smaller. Here we have incorporated the phase shift for the first time. Since inside the parenthesis, y = 3 * cos(2θ+ 7/5) + 1, we have a plus sign, we shift the equation horizontally to the left as now a negative number is needed to get back to zero just like the first number in our interval, -7/10 ≤ x ≤ π - 7/10.

Any point on the organish graph directly to the left of the blue graph is 7/10 units from each other. Let's check to see what happens when we add the phase shift and the vertical shift by graphing y = 3 * cos(2θ + 7/5), and y = 3 * cos(2θ +7/5) + 1.

The blue line represents y = 3 * cos(2θ); the orangish line represents y = 3 * cos(2θ + 7/5); the greenish line represents y = 3 * cos(2θ +7/5) + 1.