Graphs of trig functions
Graphs of trig functions
None
Example: Graph of cosine
Basic interpretation of the graph of the cosine function
Discussion and questions for this video
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 In the graph below, for what values of theta
 does cosine of theta equal 1, and for what values of theta
 does cosine of theta equal negative 1?
 And they very nicely graphed it for us.
 The horizontal axis is the theta axis,
 and the vertical axis is the yaxis.
 And so this is the graph of y is equal to cosine of theta.
 And it makes sense with our unit circle definition.
 And I'll just make sure that we're comfortable with that.
 Because with our unit circle definition
 let me draw ourselves a unit circle.
 And I'm just going to draw it very roughly,
 just so that we get the general idea of what's going on here.
 When theta is equal to 0, we're at this point right over here
 on the unit circle.
 Well, what's the xcoordinate of that point?
 Well, it's 1.
 And you see, when theta is equal to 0 on this graph,
 cosine of theta is equal to 1.
 When theta is equal to pi/2, we're
 at this point on the unit circle.
 And the xcoordinate is what?
 Well, the xcoordinate there is 0.
 And you see once again, when we're at pi/2,
 the xcoordinate is 0.
 So this is completely consistent with our unit circle
 definition.
 As we move in the rightward direction,
 we're moving counterclockwise around the unit circle.
 And as we move in the leftward direction,
 we're moving counter oh, sorry.
 If we move in the rightward direction,
 we're moving counterclockwise.
 And as we're moving in the leftward direction,
 along the axis in the negative angles,
 we're moving in the clockwise direction around our unit
 circle.
 So let's answer their question.
 For what values of theta does cosine of theta equal 1?
 Well, we can just read the graph right over here.
 It equals 1.
 So cosine of theta equals 1 at theta is equal to well,
 we see it right over here.
 Theta is equal to 0.
 Theta is equal to well, we've got
 to go all the way again to 2 pi.
 But then it just keeps going on and on, and it makes sense.
 Cosine of theta, the xcoordinate
 on this unit circle, equaled 1 right when we're at 0 angle.
 And we had to go all the way around the circle
 to get back to that point, 2 pi radians.
 But then it'll be again when we get to 4 pi radians, and then
 6 pi radians.
 So 2 pi, 4 pi, 6 pi, and I guess you could see the pattern here.
 We're going to keep hitting cosine of theta
 equals 1 every 2 pi.
 So you can really kind of view this as every multiple of 2 pi.
 2 pi n where n is an integer.
 And that applies also for negative values.
 If you're going the other way around,
 we don't get back until we get to negative 2 pi.
 Notice, we were at 0.
 And then the next time we're at 1 again
 is at negative 2 pi and then negative 4 pi,
 and then over and over and over again.
 But this applies.
 If n is an integer, n could be a negative number.
 And so we get to all of the negative values of theta,
 where cosine of theta is equal to 1.
 Now let's think about when cosine of theta
 is equal to negative 1.
 So cosine of theta is equal to negative 1
 at theta is equal to well, we can just
 look at this graph right over here.
 Well, when theta is equal to pi.
 And let's see, it kind of goes off this graph,
 but this graph would keep going like this.
 And you'd see it would also be at 3 pi.
 And you can visualize it over here.
 Cosine of theta is equal to negative 1
 when we're at this point on the unit circle.
 So that happens when we get to pi radians.
 And then it won't happen again until we
 get to 2 pi, 3 pi radians.
 And it won't happen again until we add another 2 pi,
 until we make one entire revolution.
 So then that's going to be 5 pi radians.
 And you can keep going on and on and on.
 And that's also true in the negative direction.
 So if we take 2 pi away from this, so if we were here.
 And if we go all the way around back to negative pi,
 it should also be the case.
 And you actually see it right over here on the graph.
 So you could think about this as 2 pi n plus pi.
 Or you could view it as 2n plus 1, or 2n plus 1 times pi
 where n is an integer.
 Let me write that a little bit neater.
 At every one of those points, for every one of these thetas,
 cosine of theta is going to keep hitting negative 1 over
 and over again.
 And you see it.
 It goes from one bottom or you can [? call it ?] valley
 to the next valley.
 It takes 2 pi to get to the next valley.
 And that was also the same thing for the peaks.
 It took 2 pi to go from the top of one hill
 to the top of the next.
 And then 2 pi again to the top of the hill after that.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?

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This discussion area is not meant for answering homework questions.
What is exactly happening at1:20 and why is 'y' taken as equal to cosQ?
im learning this myself and this is what i have so far.
"y" is taken as the equal to cos(theta) because that is what you are measuring. remember the "x" axis is actually the "theta" axis. you you are graphing where cos(theta) is when theta is at a specified value. so xaxis=theta and yaxis=cos(theta). if you were trying to find the sin(theta), this would be a different graph, then that would make yaxis=sin(theta).
i was at first confused from the fact that the coordinates were supposed to be (cos, sin), but then realized that this only applied to the unit circle.
at 1:20 he is showing the direction you should travel on the unit circle to get the correct coordinates for the graph. if you move the wrong direction, you aren't getting the correct numbers. you can really tell when you start graphing.
"y" is taken as the equal to cos(theta) because that is what you are measuring. remember the "x" axis is actually the "theta" axis. you you are graphing where cos(theta) is when theta is at a specified value. so xaxis=theta and yaxis=cos(theta). if you were trying to find the sin(theta), this would be a different graph, then that would make yaxis=sin(theta).
i was at first confused from the fact that the coordinates were supposed to be (cos, sin), but then realized that this only applied to the unit circle.
at 1:20 he is showing the direction you should travel on the unit circle to get the correct coordinates for the graph. if you move the wrong direction, you aren't getting the correct numbers. you can really tell when you start graphing.
y is not equal to theta the function you are graphing is y=costheta
I was confused too! Thanks to the people that
responded above!!
responded above!!
When you look at the one unit circle, then the x coordinate of the point is cos of theta. In the graph there, cos of theta is a function, and is named y=cos of theta. That <b>does not</b> mean that in the circle the y coordinate is cosine!
The x values correspond with measures of the angles in the circle. As you go around the circle type in your calc cosine90, cosine180, cosine270, cosine360, and you will get the yvalues. At each degree there is a corresponding x value. 90 degrees it is pi/2
180 degrees = pi,
270 degrees =3pi/2
360 degrees = 2pi
180 degrees = pi,
270 degrees =3pi/2
360 degrees = 2pi
cos(theta) = Adjacent over hypoteneuse
Since we are working with a unit circle, it is given that the hypoteneuse is 1, therefore
cos(theta) = adjacent over 1, shortened to just adjacent.
The adjacent side as long as the hypoteneuse, running parallel to the xaxis, resulting in a termination point of (1,0)
Since we are working with a unit circle, it is given that the hypoteneuse is 1, therefore
cos(theta) = adjacent over 1, shortened to just adjacent.
The adjacent side as long as the hypoteneuse, running parallel to the xaxis, resulting in a termination point of (1,0)
y is taken as equal to cosQ because it is the dependent variable, that is it depends on the value of theta.
Is there some sort of intuition as to why cos is the x of the unit circle and sin is the y?
FishHead,
Well, yes, it's all based on how we define the angle that we are interested in when we are using the unit circle (most people call that angle θ, theta). Just as a convention, we define θ as the angle from the X axis, to the line that goes from the origin to the point on the circle we are interested in. So when we talk about the point on the unit circle at 45 degrees, ( or pi/4 radians), that's 45 degrees from the positive X axis.
If we have some point on the unit circle, and we draw a line from the origin to that point, that gives us 1 leg of a triangle. If we use the X axis as the second leg of the triangle, and then draw a line that goes through the point on the circle to the x axis (and we make sure the line is perpendicular to the x axis), that gives us the third leg of our right triangle. If you do this, you'll see that the angle, θ, that I was talking about before is one of the three angles of this right triangle. It is the angle at the point of the triangle that touches the origin. So if we take the cosine of that angle, that is equal to the adjacent side of that triangle divided by the hypotenuse. But notice that the adjacent side of the triangle is just the x coordinate for the original point on the unit circle that we are interested in, and the hypotenuse of the circle is just 1. So the adjacent side over the hypotenuse is just the x coordinate of the point on the circle divided by 1.
Similarly, the sign of the angle θ is the opposite side of the triangle that we drew, divided by the hypotenuse. But this is just the y coordinate of the point on the circle, divided by 1… which is just the ycoordinate of the point on the circle.
If this doesn't make sense, it may help to draw a pair of x and y axes, and draw a circle on the paper centered on the origin with a radius of 1. Then, pick a point on the circle (and I would pick a point in quadrant 1 until you get comfortable with the concepts), and draw the appropriate right triangle.
Well, yes, it's all based on how we define the angle that we are interested in when we are using the unit circle (most people call that angle θ, theta). Just as a convention, we define θ as the angle from the X axis, to the line that goes from the origin to the point on the circle we are interested in. So when we talk about the point on the unit circle at 45 degrees, ( or pi/4 radians), that's 45 degrees from the positive X axis.
If we have some point on the unit circle, and we draw a line from the origin to that point, that gives us 1 leg of a triangle. If we use the X axis as the second leg of the triangle, and then draw a line that goes through the point on the circle to the x axis (and we make sure the line is perpendicular to the x axis), that gives us the third leg of our right triangle. If you do this, you'll see that the angle, θ, that I was talking about before is one of the three angles of this right triangle. It is the angle at the point of the triangle that touches the origin. So if we take the cosine of that angle, that is equal to the adjacent side of that triangle divided by the hypotenuse. But notice that the adjacent side of the triangle is just the x coordinate for the original point on the unit circle that we are interested in, and the hypotenuse of the circle is just 1. So the adjacent side over the hypotenuse is just the x coordinate of the point on the circle divided by 1.
Similarly, the sign of the angle θ is the opposite side of the triangle that we drew, divided by the hypotenuse. But this is just the y coordinate of the point on the circle, divided by 1… which is just the ycoordinate of the point on the circle.
If this doesn't make sense, it may help to draw a pair of x and y axes, and draw a circle on the paper centered on the origin with a radius of 1. Then, pick a point on the circle (and I would pick a point in quadrant 1 until you get comfortable with the concepts), and draw the appropriate right triangle.
It is applied to just about every function that is graphed, you put the input on the x axis, then the output on the y axis, and if it is your goal, you then put in lines to connect your points. It will also work in reverse.
At 2:05 about it states that when cos= 1 then theta is 0; I'm confused, I thought cos equalled one at 2π.
2pi and 0 are the same thing. 2pi is once all the way around the circle.
Cant the value of theta that give cos(theta) = 1 be :
π , 3π , 5π and so on ? As cos(theta) also becomes 1 as we go counterclockwise in the unit circle .
So can't it be represented as 2πn + π ?
π , 3π , 5π and so on ? As cos(theta) also becomes 1 as we go counterclockwise in the unit circle .
So can't it be represented as 2πn + π ?
Your answer and Sal's answer are actually the same because n can be any integer. This, when n is negative for your expression but positive (with the same absolute value) for Sal's expression, both expressions will have a value of 1 when the cosine is taken of them. However, your answer isn't "uniform" because the multiple of n must be positive and the added constant must be in the interval [0, 2*pi) for the answer to be "uniform."
Example:
2πn+π for n=1 and 2πn+π for n=1
3π and 3π
Take the cosine of both expressions.
1 and 1
I hope this helps!
Example:
2πn+π for n=1 and 2πn+π for n=1
3π and 3π
Take the cosine of both expressions.
1 and 1
I hope this helps!
How do you calculate the cos/sin/tan/other trigonometric functions for imaginary numbers?
You may not be able to. I don't think theta can be imaginary.
Doesn't cos=1 at 2pi also?
And also, doesn't the question say just "in the graph below"? Why doesn't he just end the answer with cos(theta)= 2pi, 0, 2pi ?
And also, doesn't the question say just "in the graph below"? Why doesn't he just end the answer with cos(theta)= 2pi, 0, 2pi ?
Yes, it continues on that pattern toward an infinity of radians, both positive and negative. I cannot speak for why Sal did not stop at the narrow definition of the question, but at that point in the video, he was showing that the pattern continues on and on, and how to find out how it continues. He gave us the general form for all values when cos(theta) =1 and 1, which was also a *bonus result* to help us calculate for ourselves, but was not part of the original question. It is true that on an exam you would probably want to stick to the exact answer as requested.
Is it ever possible for a graph of COS or SIN to have a value that is undefined
Sec(x) is undefined when cos(x) is zero, because sec(x) is 1/cos(x). You can't divided by zero. Same thing with cosecant. Tangent is undefined at pi/2 and 3pi/2, because the line through the origin becomes vertical at those points.
so the graphs of CSC, SEC, and COT will have undefined values at
CSC 0 radians, pi
CSC 0 radians, pi
It's a greek letter which often represents an angle.Like a variable
Theta is a greek letter that represents an angle, much like alpha or beta.
It's pretty much x but for angles.
It's a greek letter to called an angle
It is a greek letter which sometimes represents an angle sort of like x represents an angle.
It's an angle measure (degrees or radians)
4:40 Couldn't you also say pi*n, for n of all odd integers?
Yes, but it's a bit clunkier. Sal's solution is more conventional. We tend to like to define n as a member of the integers (a wellknown set of numbers) and then write `theta=(2n+1)pi`, which takes care of the "odd numbered multiples" of pi we're after.
at 3:50, why is it that it goes pi, 3 pi, 5 pi for cos (1) rather than o, 2, 4 pi...?
The cosine is just the xvalue of a point on the unit circle. At 0, 2pi, 4pi, etc, the xvalue is 1. But at pi, 3pi, 5pi, etc, the xvalue is 1.
What would you do if you are trying to find the zeros of a function, for example 3 cosx?
The graph of 3 cos(x) has the same zeros as the regular graph of cos(x), despite that 3 cos(x) is vertically stretched by a factor of 3.
At 0:58 why does Cos π/2 equal to 0?
π is equal to 180 degrees. Therefore 2π equals 360 degrees, which is along the x axis. So looking at the unit circle Cos is equal to X.
How would I find the natural domain of trigonometric functions such as: 3/(2cosx)?
All I'm interested in is the method/concept involved.
All I'm interested in is the method/concept involved.
Domain is always the values of "x" where a you get a real solution.
So ask yourself "what makes this function unsolvable?"
The most common culprits are square roots of negatives and dividing by zero (though others do exist). In this case, is there anything that makes the denominator equal to zero? If so, you would say the domain is all x EXCEPT for x="whatever those are."
So ask yourself "what makes this function unsolvable?"
The most common culprits are square roots of negatives and dividing by zero (though others do exist). In this case, is there anything that makes the denominator equal to zero? If so, you would say the domain is all x EXCEPT for x="whatever those are."
Why does Sal use the "y" for the vertical axis for the sinusoidal graph? I think it would be less confusing if Sal plotted "X" for the vertical axis for the sinusoidal function. I think the confusion for me was that you look and at the sinusoidal wave and see "y" when the cosine for a unit circle plots the Xcordinate value. You think you should be looking at the ycoordinate axis on the Cartesian coordiante graph, but Cos theta = Adjacent/Hypotenuse where the Hypotenuse is "1." Therefore, you should be looking at the Xcoordinate for the various cosine angles.
How do I find "b" of a cosine graph? I do not understand the cyles and how to count them.
B is the coefficient in front of the X value in the parentheses.
To find it you take the period of the graph (the time it takes to repeat itself) and set that equal to 2B. This works because the period in the parent graph [f(x)=Cos(x)] the coefficient in front of your x value (or B) is 1, yet the period is 2. This means the period is always twice as much as your B value.
To find it you take the period of the graph (the time it takes to repeat itself) and set that equal to 2B. This works because the period in the parent graph [f(x)=Cos(x)] the coefficient in front of your x value (or B) is 1, yet the period is 2. This means the period is always twice as much as your B value.
By "b", do you mean the Y intercept? If so, look at where the graph intercepts the Y axis. In this case, it intercepts at point (0,1), so b = (0,1).
when exactly will n be represented in (2pi) n
I think what you mean by your question is, "Why do they put the n there and what does it mean?"
If not, could you please clarify? Thanks. :)
Meanwhile, I will answer what I believe you are asking.
The n represents all real numbers. These are also known as "all reals" or just as "R".
So it can be a 5, a 50, a billion... whatever real number you please.
So why do we put it there? Well, instead of listing all the infinite multiples of 2pi, we put n and say that n = all reals.
Okay. So what does it mean? It tells us that the answer is any multiple of 2pi. In trig, this means that every time we make a full circle, we have the wanted answer.
Since your asking this question on the "graph of cosine" video, I assume you wanted to know this question due to a cosine problem.
And what n(2pi) means when we are talking about cosine is that every time we make a full circle, we are going to get 0(on a unit circle).
Now I have a question for you: Are you sure it said n(2pi), not n(pi)? Because you get the same answer for cosine in both answers....
Hope this helps. :)
Sylvia.
If not, could you please clarify? Thanks. :)
Meanwhile, I will answer what I believe you are asking.
The n represents all real numbers. These are also known as "all reals" or just as "R".
So it can be a 5, a 50, a billion... whatever real number you please.
So why do we put it there? Well, instead of listing all the infinite multiples of 2pi, we put n and say that n = all reals.
Okay. So what does it mean? It tells us that the answer is any multiple of 2pi. In trig, this means that every time we make a full circle, we have the wanted answer.
Since your asking this question on the "graph of cosine" video, I assume you wanted to know this question due to a cosine problem.
And what n(2pi) means when we are talking about cosine is that every time we make a full circle, we are going to get 0(on a unit circle).
Now I have a question for you: Are you sure it said n(2pi), not n(pi)? Because you get the same answer for cosine in both answers....
Hope this helps. :)
Sylvia.
How do u know when what part cuts the line and where i have a bad understanding of this part
No offense, sir, but I have a bad understanding of this question.
Hi, I'm not quite understanding why cos(pi)=cos(pi). cos(pi) gives me negative one, so shoulden't cos (pi) give me the reverse of positive one? Thanks so much.
Well, look at the graph of cosine. It is even, which means that it is symmetric around the yaxis. If something is symmetric is around the yaxis, then f(x)=f(x). So, cos(pi)=cos(pi)= 1.
How would one find the xintercepts of a trigonometric function given the equation and not the graph?
Well, basically, if you want to know the xintercepts, you want to know *when the value of the function is equal to zero*, because the value of f(theta) is the vertical axis in this case.
So, *it depends on the function* from that point.
When does `sin(theta) = zero`? According to SohCahToa, sine is opposite over hypotenuse, and sine(theta)'s value is zero when the opposite side is zero. After a while you will have this figured out, but until you do, o figure this out, draw a little picture of the unit circle like Sal did at 0:30
Now, when is the opposite side (of the unit circle triangle) equal to zero? It happens twice on the way around the unit circle: when the hypotenuse is squashed against the xaxis. This occurs when `theta = 0` and also when `theta = pi` (180 degrees from the initial side)
But we don't stop there, because the circle is continuousthe angle can keep on increasing, going around and around and around past the horizontal axis every time it passes through another pi's worth of revolution.
So sin(theta) =0 every additional pi, and if you go the other way around the unit circle, it has a value of zero at every additional 1pi revolutions. So for what you need to know, on a graph of the sine function graphed against the angle in radians, the xintercepts will occur at every positive and negative whole number multiple of pi continuing out *forever*, in addition to when the angle measures 0 radians ( degrees)
Sal just did the cos(theta) above, and the xintercepts for cosine are staggered from sine by 90 degrees (pi/2 radians). Now we care about Coh, or opposite over adjacent on the unit circle. Again, you will have an xintercept whenever the value of cosine equals 0.
This happens at pi/2, which is 90 degrees, because that is when the length of the adjacent side of the unit circle triangle becomes zero. It also happens at every additional added 180 degrees beyond pi/2, for example at 3/2 pi radians, 5/2 pi radians (270 degrees and 450 degrees and on and on in the forward direction). If you rotate the angle backwards, the value of cos(theta) is zero as you subtract pi (180 degrees) from pi/2 radians. So cos(theta) =0 and you have xintercepts every negative rotation that ends with the value of 0: 3/2 pi, 5/2 pi and on and on.
The graph of tangent is strangelooking because it runs off the graph at regular intervals, looking like a row of snakes. Tangent is not defined when the angle is pi/2 because tangent equals opposite over adjacent, and at 90 degrees, the adjacent side is zero. It does have an `xintercept every time sine equals zero and cos equals one (or negative one)`, which happens at theta = 0 and theta = pi and theta = 2pi and also, going the other direction, at theta = 1pi radians and theta = 2 pi radians.
Cotangent looks like tangent, only the snakes wiggle the other direction and the xintercepts occur at offset from the intercepts of tangent. Cotangent(theta) equals zero when cosine equals zero, in other word when theta = pi/2, 3/2 pi, 5/2 pi, and also, pi/2, 3/2 pi and so on.
Well, that is a long answer and I have not gotten to cosecant and secant, but never fear. The process is the same.The surprising result for secant is that it is the inverse function for cosine, and its value *never crosses the xaxis*. Same thing for cosecant, the inverse function for sine. So the quick answer is there are `no xintercepts for secant and cosecant.`
So, *it depends on the function* from that point.
When does `sin(theta) = zero`? According to SohCahToa, sine is opposite over hypotenuse, and sine(theta)'s value is zero when the opposite side is zero. After a while you will have this figured out, but until you do, o figure this out, draw a little picture of the unit circle like Sal did at 0:30
Now, when is the opposite side (of the unit circle triangle) equal to zero? It happens twice on the way around the unit circle: when the hypotenuse is squashed against the xaxis. This occurs when `theta = 0` and also when `theta = pi` (180 degrees from the initial side)
But we don't stop there, because the circle is continuousthe angle can keep on increasing, going around and around and around past the horizontal axis every time it passes through another pi's worth of revolution.
So sin(theta) =0 every additional pi, and if you go the other way around the unit circle, it has a value of zero at every additional 1pi revolutions. So for what you need to know, on a graph of the sine function graphed against the angle in radians, the xintercepts will occur at every positive and negative whole number multiple of pi continuing out *forever*, in addition to when the angle measures 0 radians ( degrees)
Sal just did the cos(theta) above, and the xintercepts for cosine are staggered from sine by 90 degrees (pi/2 radians). Now we care about Coh, or opposite over adjacent on the unit circle. Again, you will have an xintercept whenever the value of cosine equals 0.
This happens at pi/2, which is 90 degrees, because that is when the length of the adjacent side of the unit circle triangle becomes zero. It also happens at every additional added 180 degrees beyond pi/2, for example at 3/2 pi radians, 5/2 pi radians (270 degrees and 450 degrees and on and on in the forward direction). If you rotate the angle backwards, the value of cos(theta) is zero as you subtract pi (180 degrees) from pi/2 radians. So cos(theta) =0 and you have xintercepts every negative rotation that ends with the value of 0: 3/2 pi, 5/2 pi and on and on.
The graph of tangent is strangelooking because it runs off the graph at regular intervals, looking like a row of snakes. Tangent is not defined when the angle is pi/2 because tangent equals opposite over adjacent, and at 90 degrees, the adjacent side is zero. It does have an `xintercept every time sine equals zero and cos equals one (or negative one)`, which happens at theta = 0 and theta = pi and theta = 2pi and also, going the other direction, at theta = 1pi radians and theta = 2 pi radians.
Cotangent looks like tangent, only the snakes wiggle the other direction and the xintercepts occur at offset from the intercepts of tangent. Cotangent(theta) equals zero when cosine equals zero, in other word when theta = pi/2, 3/2 pi, 5/2 pi, and also, pi/2, 3/2 pi and so on.
Well, that is a long answer and I have not gotten to cosecant and secant, but never fear. The process is the same.The surprising result for secant is that it is the inverse function for cosine, and its value *never crosses the xaxis*. Same thing for cosecant, the inverse function for sine. So the quick answer is there are `no xintercepts for secant and cosecant.`
At 1:06, he says the xcoordinate is at 0. However, it looks like x is at 1 and the ycoordinate is at 0. Am I missing something here?
On the unit circle, cos(1/2pi) = 0. Since on the graph y = cos(x) and x = theta, it looks like x isn't equal to 0, but what he's graphing is theta. On the unit circle, though, x = 0.
he means to say when xcoordinate is 0 then y=1(pointing that one) and at ycoordinate is 0 x=1
I had it. At least, I thought I had it. I've understood everything up to this video. I cannot understand why, when he starts with theta = 0, and states that at this point the X coord is 1, does he then place the point at Y = 1? It seems counterintuitive to everything I have ever learned about the X and Y axis.
The unit circle and cosine function are separate graphs. On the unit circle the cosine of any theta is given by the xcoordinate of a point on the circle. But on the graph of the cosine function the value of the cosine is placed on the yaxis instead. That's why he puts the point at y=1 and not x=1. It is a bit confusing at first.
Well, you think about x as theta and y as its cos and think about the unit circle, it comes out easily... try to be objective, because it works just like normal x and y.
Do you have a video on how to graph a tangent function? I can't seem to find one on this website. Thanks!
Is this what you're looking for?
https://www.khanacademy.org/math/trigonometry/trigfunctiongraphs/trig_graphs_tutorial/v/tangentgraph
https://www.khanacademy.org/math/trigonometry/trigfunctiongraphs/trig_graphs_tutorial/v/tangentgraph
at 4:44 , can't we just multiply it with and odd integer to find cos thetha=1?
at 2:45 how come N cannot be a decimal or fraction because an integer cannot be a fraction or decimal and Mr.Khan describes N as a integer which means N cannot be a decimal or fraction I think Mr.Khan meant a rational or irrational number? What do you think?
He stated it correctly, n MUST be an integer. That has to do with the period of the cosine function. The pattern of the cosine function repeats its self every 2π. So the n represents how many periods you go through and only a complete period is enough. So, n must be an integer.
In other words:
cos (x) = cos [ x + 2π (any integer) ]
In other words:
cos (x) = cos [ x + 2π (any integer) ]
as in cos theta there is interval of 2pi for +ve n 2n+1 for ve is it the same for sin also??
@3:14pm
@3:14pm
How does theta = 0 gives the value of co(theta) = 1 ?
It is not co(theta) it is cos(theta) cos is the abbreviation for the trignometric ratio cosine. Cos(0) = 1 where theta = 0
Theta 0 means that it is a triangle with a 0 degree triangle like http://math.colgate.edu/~kellen/interspace/0degrees.gif
So the hypotenuse and adjacent both are the same while the altitude is 0. Cos(theta)= adjacent / hypotenuse = 1
Theta 0 means that it is a triangle with a 0 degree triangle like http://math.colgate.edu/~kellen/interspace/0degrees.gif
So the hypotenuse and adjacent both are the same while the altitude is 0. Cos(theta)= adjacent / hypotenuse = 1
When theta equals zero, the hypotenuse of the triangle is squashed against the xaxis, so the adjacent side of the "triangle" is the same as the hypotenuse: they both measure 1 unit.
So the Cah in Soh Cah Toa says the value of cosine is adjacent over hypotenuse =1/1 = 1
This squash when the hypotenuse is as long as the adjacent side also happens when theta equals 1 pi (180 degrees), only the measure then is 1/1 = 1 because the axis is in the negative direction.
And so on, every time you rotate again around the circle and hit a multiple of pi and 2 pi.
So the Cah in Soh Cah Toa says the value of cosine is adjacent over hypotenuse =1/1 = 1
This squash when the hypotenuse is as long as the adjacent side also happens when theta equals 1 pi (180 degrees), only the measure then is 1/1 = 1 because the axis is in the negative direction.
And so on, every time you rotate again around the circle and hit a multiple of pi and 2 pi.
what is an asymptote ?
Thank you.
Thank you.
Hope this can help: http://en.wikipedia.org/wiki/Asymptote (or this definition from Google: a line that continually approaches a given curve but does not meet it at any finite distance)
It is a part of the graph where an output is *impossible* (i.e. 1/0 because this is domain). It's shown as a dotted line on graphs such as sec, csc, tan, and cot. The tangent of pi/2 (0,1) (sin/cos = tan) is _undefined_ so on the graph there is an *asymptote*, such as those on named graphs.
_Hope this helped._
_Hope this helped._
"where N is an integ... could probably write that a bit neater" writes it messier.
At 4:15, I'm a little bit confused. Can someone explain? Thanks!
What does this actually mean scientifically speaking...errr... well
That means that cos( π ) = 1. We deduce that because cos( x ) = cos( x + 2π ),
and cos( π ) = cos( π + 2π ) = cos( π ) = 1.
and cos( π ) = cos( π + 2π ) = cos( π ) = 1.
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