Main content

## Algebra 2 (Eureka Math/EngageNY)

### Course: Algebra 2 (Eureka Math/EngageNY) > Unit 2

Lesson 2: Topic A: Lessons 8-10: Trigonometric graphs and radians# Graph of y=sin(x)

The graph of y=sin(x) is like a wave that forever oscillates between -1 and 1, in a shape that repeats itself every 2π units. Specifically, this means that the domain of sin(x) is all real numbers, and the range is [-1,1]. See how we find the graph of y=sin(x) using the unit-circle definition of sin(x). Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- Why are imaginary numbers not included in the domain? What prevents them from 'working'?(63 votes)
- Because the domain (and range) must be within the cartesian plane (the x-y graph). Each axis on the cartesian plane contains all real numbers. Imaginary numbers exist "outside" of real numbers and therefore are not found in the cartesian plane. Numbers that can't be graphed are outside the domain and range. This illustration may help: http://www.icoachmath.com/image_md/Real%20Numbers1.gif(36 votes)

- can we get a sine curve from cosine curve,and vice versa?(19 votes)
- Yes!

sin x = cos(π/2 - x) = cos(x - π/2)

cos x = sin(π/2 - x) = -sin(x - π/2)(36 votes)

- What would the end behavior of the sine function be?

(Is there just not any end behavior since it's periodic? Or is there a special term to describe it?)(19 votes)- There is no end. A sine wave will continue into infinity, unless you restrict its domain.(29 votes)

- Why does he only input angles in multiples of 90 degrees?(7 votes)
- because it's easier and he already knows what the graph looks like, so it would not matter what angles he chose. In fact, he could have picked and number for theta and the point would lie somewhere on the wavy line. Therefore, he chose the easy ones.(23 votes)

- Why would Pi over 2 be 90 degrees instead of 180 degrees?2:17(4 votes)
- Because the circumference of a circle is
`2πr`

. Using the unit circle definition this would mean the circumference is`2π(1) or simply 2π`

. So half a circle is`π`

and a quarter circle, which would have angle of 90° is`2π/4 or simply π/2`

. You bring up a good point though about how it's a bit confusing, and Sal touches on that in this video about Tau over Pi. https://www.khanacademy.org/math/trigonometry/unit-circle-trig-func/long-live-tau/v/tau-versus-pi(20 votes)

- sin (90 degree) = opp/hyp . I agree that opp is 1, but what is the hypotenus ?(1 vote)
- Yes. On the unit circle, the hypotenuse is always the radius, 1. We define the sine of the angle as the y coordinate, so at 90 degrees our coordinates are (0,1) and it equals the radius 1. Thus we get sin 90 = 1 and cos 90 is the x coordinate so it is thus defined as zero.(11 votes)

- Can we say that the domain of a sine function is minus infinity to plus infinity? Is that the same as saying the domain is 'all Real numbers' (at8:11)?(4 votes)
- Yes, the domain of a sine function is from negative to positive infinity. The domain is written as (-∞,∞) in interval notion. It it the same thing as the domain is all real numbers.(5 votes)

- I'm pretty new to this so can some help me on what a "unit circle" is?(1 vote)
- The unit circle is used to help you find the exact values of trig functions of special angles (0°, 30°, 45°, 60°, 90° or their radian counterparts) and the multiples of those special angles. The circle uses the idea of symmetry to find the coordinates at which multiples of the special angles intersect the circle. The coordinates themselves are found using the ratios in a 30°-60°-90° triangle or a 45°-45°-90° triangle when the hypotenuse (radius of the circle) is 1.

http://www.timvandevall.com/wp-content/uploads/unit-circle-chart.png(8 votes)

- Without using calculator/four figure table,find cos b if sin b is 0.8(2 votes)
- You can find this with the Pythagorean identity, sin²x+cos²x=1. We know sin(b)=0.8, so sin²(b)=0.64, and

we have 0.64+cos²(b)=1

cos²(b)=0.36

cos(b)=±0.6

This is all we can determine without more information about which quadrant b lies in.(8 votes)

- how do you solve csc[fo](30 degrees)(3 votes)
- Hey! Do you mean the cosecant of 30 degrees? Cosecant is just the reciprocal of sine. Find the sine of 30 degrees, which is 0.5, and then take the reciprocal of that value, which is 2.(4 votes)

## Video transcript

- [Narrator] We're asked
what are the domain and range of the sine function? So to think about that,
let's actually draw the sine function out,
and what I have here, on the left hand side right
over here I've got a unit circle, and I can, let me
truncate this a little bit. I don't need that space right there, so let me clear that out. So I have a unit circle
on the left hand side right over here, and I'm
gonna use that to figure out the values of sine of
theta for a given theta. So, on the unit circle
this is X, and this is Y, or you could even do this, as the, well, we can just use X or Y,
and so for a given theta we can see where that angle
entered to the terminal side of the angle intersects the
unit circle, and then those, the Y coordinate of that point is going to be sine of theta. And over here I'm going to graph. Still Y in the vertical axis, but I'm gonna graph the graph
of Y is equal to sine of theta. Y is equal to sine of theta,
and on the horizontal axis I'm not gonna graph X,
but I'm gonna graph theta. You can do theta as the
independent variable here, and it's gonna be theta is
going to be in radians. So we're essentially going
to pick a bunch of thetas and then come up with
what sine of theta is, and then graph it. So let's set up a little
bit of a table here. Let's set up a little bit of a table. And so, over here I have theta, and over here we're going to figure out what sine of theta is, and we could do a bunch of theta values. We could start, we could start,
let's say we start at zero. Let's say we start at
theta is equal to zero. What is sine of theta gonna be? Well, when the angle is zero we intersect the unit
circle right over there. The Y coordinate of this is still zero. This is the point, this is
the point one comma zero. The Y coordinate is zero,
so sine of theta is zero. We could say sine of
zero is equal to zero. Sine of zero is equal to zero. Now let's try theta is
equal to pi over two. Theta is equal to pi over two. I'm just doing the ones that
are really easy to figure out. So if theta is equal to pi over two, that's the same thing
as a 90 degree angle. So, the terminal side is going
to be right along the Y axis just like that, and where it
intersects, where it intersects the unit circle is right over
here, and what point is that? Well that's the point zero comma one. So what is sine of pi over two? Well sine of pi over two is just the Y coordinate right over here. It is one. Sine of pi over two is one. Let's keep going and you might
see a little pattern here. We're just going more and
more around the circle. So let's think about what's, what, what happens when theta is equal to pi. When theta is equal to
pi, what is sine of pi? Well, we intersect the unit
circle right over there. That coordinate is negative one, zero. Sine is the Y coordinate, so this right over here is sine of pi. Sine of pi is zero. Let's go to three pi over two. Three pi over two, well now
we've gone three quarters of the way around, around the circle. We intersect the terminal
side of the angle intersects the unit circle right over here, and so based on that what is
sine of three pi over two? Well, this point right over
here is the point negative, we gotta be careful, is
zero, is zero negative one. The sine of theta is the
same thing as a Y coordinate if the Y coordinate is
the sine of theta, so theta, when theta's pi over two sine of theta, or when theta's three pi over two sine of theta is equal to negative one. And let's come full circle. Let's come full circle here. So let's go all the way
to theta equaling two pi. Let me do a color, hey, I'll
just use the yellow here. What happens when theta is equal to two pi? Well then we've gone all
the way around the circle, and we are back to where we started, and the Y coordinate is zero, so sine of two pi is once again zero, and if we were to keep going around, we're gonna start seeing as we
keep incrementing the angle, we're gonna start seeing the
same pattern emerge again. Well, let's try to graph this. So when theta is equal to
zero, sine of theta is zero. When theta is equal to pi
over two, when theta is equal to pi over two, pi over
two, sine of theta is one. So, we'll use the same scale. So sine of theta, sine
of theta is equal to one. This is, I'll just make this, this is one on this
axis, and on that axis. So we can maybe see a little
bit of a parallel here. When theta is equal to
pi, sine of theta is zero. So when theta is equal to
pi, sine of theta is zero. So we go back right over there. When theta is equal to three pi over two, so that would be right over
here, three pi over two, sine of theta is negative one. So this is negative one over here. I'll do the same scale over here. I'll make this negative. I'll make, let me make down a little bit, I'll make this negative one, and so, sine of theta is negative one. And then, when theta is two
pi, sine of theta is zero. So when theta is two pi, two
pi, sine of theta is zero. And so we can connect the dots. You could try other points in between and you get something, you get a graph that looks something like this. It looks something like this. My best attempt at drawing it freehand. It looks something, something like this. There's a reason why curves
that look like this are called sinusoids, because they're
the graph of a sine function. So this like, just like this, but that's not the entire graph. We could keep going. We could go, we could
add another pi over two. If you added another pi over
two, so if you go to two pi, and then you add another pi over two. So you could view this
as two and a half pi, or however you wanna think about it, then you're gonna go back over here. So then you're gonna get back to sine of theta being equal to one. So you're gonna go back to
this point right over here, and you could keep going. You go another pi over
two, you're gonna go back to this point, and you're
gonna be over here, and so the curve, the curve
or the function sine of theta is really defined for any theta value, any real theta value that you choose. So any theta value. Well, what about negatives? I mean obviously I agree,
as you keep increasing theta like this we just keep going
around and around the circle and this pattern kind of emerges, but what happens when we go
in the negative direction? Well, let's try it out. What happens if we were to
take, if we were to take negative pi over two? So let me do that. So negative pi over two, well,
that's going right over here, and so we intersect the unit
circle right over there. The Y coordinate is negative one. So sine of negative pi
over two is negative one, and we see that it just continues. It just continues. So sine of theta is defined
for any positive, negative, or any theta, positive or negative, non negative, zero, anything. So it's defined for anything. So, let's go back to the question. So I could just keep drawing
this function on and on and on. So let's go back to the question. What is the domain, what is the domain? What is the domain of
sine of the sine function? And just as a reminder, the
domain are all of the inputs over which the function is defined, or all of the valid
inputs into the function that the function will actually
spit out a valid answer. So what is the domain
of the sine function? Well, we already saw. We can put in any theta here. So you could say the
domain, the domain is all, all real numbers, all
real, all real numbers. Now, what about the range? What about the range? Well just as a review,
the range is sometimes in more technical math
classes called the image. It's the set of all the values that the function can actually take on. Well what is that set? What is the range here? What is all the values that Y equals sine of theta could actually take on? Well, we see that it keeps
going between positive one, it keeps going between positive
one and then to negative one and then back to positive
one and then negative one. It takes on all the values in between. So you see that sine of
theta, sine of theta is always going to be less than or equal to one, and it's always going to be greater than or equal to negative one. So you could say that
the range of sine of theta is the set of all numbers
between negative one and positive one, and it
includes negative one and one, and that's why we put brackets
here instead of parentheses.