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## Algebra 2

### Course: Algebra 2>Unit 11

Lesson 8: Graphing sinusoidal functions

# Example: Graphing y=3⋅sin(½⋅x)-2

Sal graphs y=3⋅sin(½⋅x)-2 by thinking about the graph of y=sin(x) and analyzing how the graph (including the midline, amplitude, and period) changes as we perform function transformations to get from y=sin(x) to y=3⋅sin(½⋅x)-2. Created by Sal Khan.

## Want to join the conversation?

• For the practice problems, I don't understand why for some of the graphs, the extremun's x-coordinate is 0, and for other graphs, the midline's x-coordinate is 0.
• Oh, I figured it out! If it’s cos - the extremum point’s x-coordinate is c, and the y-coordinate is a + d. The midpoint’s x-coordinate is 2pi divided by b, then that whole thing divided by 4, and the y coordinate is d.
If it’s sin - the midpoint’s x-coordinate is c, and the y-coordinate is d. The extremum’s x-coordinate is 2pi divided by b, then that whole thing divided by 4, and the y-coordinate is a + d.

basically:

sine
extremum point : ((c), (a+d))
midpoint : (((2pi / b)/4), (d))

cosine
extremum point : (((2pi / b)/4), (a+d))
midpoint : ((c), (d))
• Hi, I know Sal mentioned that sin(0) is always 0, sin(pi/4) is always sqrt(2)/2.
Is that the case for all equations OR is it just for unit circles?
• It is only for unit circles, this is true because the side ratios of special right triangles are as follows -

30-60-90 ---> side opp to 30 is x
side opp to 60 is (root 3)*x
side opp to 90 is 2x

45-45-90 ---> side opp to 45 is x
side opp to 90 is (root 2)*x

In the case of the unit circle it is (root 3)/2 because the 2x (side opp to 90 in 30-60-90 triangle) in the case of that triangle is 1 (is a unit circle and the hyp is the radius) and therefore x = 1/2 using simple algebra. The other side is (root 3)*x and since x = 1/2, that simplifies to (root 3)/2. If the hyp of the triangle was not = 1 (means the circle is not a unit circle) then this would not work. This is why it is only for a unit circle.
• Why when you have an equation, if you have a coefficient on one side do you have to distribute that to the other side, but also making sure you distribute it to the entire expression on the other side
• I am not sure how this question relates to the video, it sounds like you are asking something more like 4x+3y=6. To get this in slope intercept form, subtract 4x to get 3y=-4x+6. To get rid of the coefficient of 3, we have to divide ALL TERMS by 3 to get 3/3y = 4/3 x + 6/3 or y=4/3x + 2. If you do not divide all terms on the other side, you will not get an equivalent equation. Example: suppose you say 10 = 6 + 4. If I divide by 2 (and do not divide all terms on right by 2) I might get 5 = 3+4 or 5=6+2, neither of which is correct. If I divide all terms by 2, I get 5 = 3+2 which is correct.
• How to derive `period = 2pi/k`?
• Let's look at sin(x), thinking about the unit circle definition. We complete a full cycle after we have rotated 2π radians, so the period of sin(x) is 2π.

In sin(kx), we go k times as fast as in sin(x). We can even imagine the unit circle as a racetrack, with x as time, sin(kx) as the y position of the car, and k as the speed of the car. How much time will it take for the car to go around the track once?

We already know that when k=1, the period is 2π. We can also use this:

speed = distance/time

We already talked about the speed being k. The period should just be the time elapsed when the car completes one lap. So, let's plug in speed = 1 and time = 2π, seeing what we get for distance.

1 = distance/2π
2π = distance

This makes sense, considering there are 2π radians in a circle. When we plug in k and period as we always have, we get this equation:

speed = 2π/time
k = 2π/period

If we multiply period and divide k on both sides, we end up with our final answer.

period = 2π/k
• So I have been doing some textbook problems related to this, and how come I have never encountered a problem in which b is negative? Is there a reason why this is the case?
(1 vote)
• I don't understand why Sal says that the Sin of (1/2)pi = 1 but then graphs the point at (pi, 1). Shouldn't the coordinates be (1/2pi, 1)?
(1 vote)
• Assuming you mean around , what Sal is saying is that if you are at x=π, you can test for what the y value is by plugging π in for x in the equation (but right then he's ignoring the 3 and -2). This makes (1/2)π, and the sin of that is 1. So at x=π, y=1.
• I understand how to graph a sin function when the x-axis uses increments of pi, but how do you create a basic sin function like Sal does at when the x-axis uses increments of 1, like in some of the questions in the next practice section?
(1 vote)
• Know that Period = 2pi/B
When the sin function is in the form A*sin(Bx) + C
If your "B" has pi in it, the pi will cancel from the numerator and denominator and give you a result that is NOT in the increments of pi, But when "b" is just a random number, you will get it in terms of pi.
Hope this helped!
• Sin graph Phase
shift using max and min points
(1 vote)
• Do you have to go up 8 units if it became 8*sin(1/2x)-2?
(1 vote)
• Not necessarily, the 8 you are talking about is the amplitude. So you actually go up AND down 8 from the midline.
(1 vote)
• Why does it make sense to divide 2pi by the period
(1 vote)
• 2pi is the standard period so if you have a different period, dividing by 2pi is like finding the ratio. It's the same thing as finding the decimal form of a percentage. You are using 100 as the standard and finding a ratio you can multiply with other numbers. Imagine you had 30 divided by 100, you would get 0.30. That result can now be multiplied by "x" to get 30% of "x". Similarly, (the period)/(2pi) gives you a number to multiply x by to get your desired value.
(1 vote)