Solving cos(θ)=1 and cos(θ)=-1

Sal solves the equations cos(θ)=1 and cos(θ)=-1 using the graph of y=cos(θ).

Solving cos(θ)=1 and cos(θ)=-1

Discussion and questions for this video
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 x-axis=theta and y-axis=cos(theta). if you were trying to find the sin(theta), this would be a different graph, then that would make y-axis=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.
Is there some sort of intuition as to why cos is the x of the unit circle and sin is the y?

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 y-coordinate 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.
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.
If you mean:
cos (θ²) + sin(θ²), then that is NOT equal to 1,
except for a few special angles such as θ=√(2π), θ=0 or θ= ½√(2π)

If you mean: (cos θ )² + (sin θ)² = 1
Which is usually written as: cos² (θ) + sin²( θ) = 1
Then that is true.
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 ?
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."

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!
Because sin = Opposite/ Hypotenuse and cos = Adjacent/ Hypotenuse
and hypotenuse of a triangle is always greater than to the other two sides. So the fraction of opp./hyp. or adj./hyp. can never be greater than 1
So basically, at cosine fanction we start rotating the circle anti-clockwise at y=1 and we count ...y=0 ~ x=π/2, y=-1 ~ x=π... and so on, while at sine fanction the circle starts 1 quarter to the right.
How do you calculate the cos/sin/tan/other trigonometric functions for imaginary numbers?
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 ?
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.
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 well-known 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 x-value of a point on the unit circle. At 0, 2pi, 4pi, etc, the x-value is 1. But at pi, 3pi, 5pi, etc, the x-value 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.
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.
What is happening at 4:27 when Sal states that Cos(theta) = -1 can be viewed as (2n + 1)pi? Why add the one to that function, when the previous function "2n(pi)" does the same thing?
(2n+1) where n is an integer, is the standard way of specifying any odd integer. Whereas 2n is even. So, cos θ = −1 only when θ is an odd integer multiplied by π.
How would I find the natural domain of trigonometric functions such as: 3/(2-cosx)?
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."
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 X-cordinate value. You think you should be looking at the y-co-ordinate axis on the Cartesian co-ordiante graph, but Cos theta = Adjacent/Hypotenuse where the Hypotenuse is "1." Therefore, you should be looking at the X-co-ordinate 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.
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. :)
is there any limit to what values of theta in cos(theta) can produce same values . like cos(0)=cos(360)..(because we come back to where we started)..and you can go on and on..
It's just like x. It's a variable that you can use for anything, but instead of being a member of all numbers, it's usually only integers.

In the context of this video, n is used to show that since `cos(θ) = cos(θ+2π)`, you can add/subtract any n number of 2π's to theta without changing the value of cosine.
Is there a video on how to find the phase shift and vertical shift of sine and cosine functions?
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.
Which video should I watch to learn how to graph an equation such as y=2sin4x OR y=3^x-1?
For an equation involving trig ratios like sine, cosine etc. you need to know 3 things about that equation: Its amplitude and period. Try watching "Example: Amplitude and period" under graphs of trig functions. Finding the amplitude will help you determine the height or how far it will stretch along the y-axis and the period will tell you how far it will stretch along the x-axis.
And as for the other equation which is also called a linear equation has nothing to do with trig, simplest way is to try putting arbitrary numbers to x or y. like example you equation says y = 3^x-1, well try putting y = 0, or x = 0 or y = 1, any number you will get the values of x and y which is after all the co-ordinates in a graph.
Theta is a variable just like x, y, a, b, etc.. Usually θ is used as a variable to represent an unknown angle. We get θ as well as many other symbols used in math (such as π) from the Greek alphabet.
Hey guys this isn't part of the video but is related to the concept. I was just going over some trigonometric functions (specifically sine and cosine functions) and was wondering, as sin45 = 1/root2, how is it calculated that sin45=1/root2? I hope this doesn't come across as a stupid question but it just baffles me. I don't have a maths background but it is a little confusing as they show no steps towards that answer.
The actual way that the trigonometric functions are computed requires advanced calculus to understand. But some of the easier ones can be figured out with right triangles and the Pythagorean theorem.
hola compadres el filma en espanol? me nessisita assistiance no comprende 2sin(2x-2(pi).
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 y-axis. If something is symmetric is around the y-axis, then f(x)=f(-x). So, cos(-pi)=cos(pi)= -1.
How would one find the x-intercepts of a trigonometric function given the equation and not the graph?
Well, basically, if you want to know the x-intercepts, 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 x-axis. 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 continuous--the 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 x-intercepts 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 x-intercepts 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 x-intercept 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 x-intercepts 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 strange-looking 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 `x-intercept 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 x-intercepts 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 x-axis*. Same thing for cosecant, the inverse function for sine. So the quick answer is there are `no x-intercepts for secant and cosecant.`
No, because n is an integer, which includes the negative integers as well.
2πn, when n=-1 is -2π.
At 1:06, he says the x-coordinate is at 0. However, it looks like x is at 1 and the y-coordinate 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.
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 x-coordinate of a point on the circle. But on the graph of the cosine function the value of the cosine is placed on the y-axis instead. That's why he puts the point at y=1 and not x=1. It is a bit confusing at first.
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) ]
as in cos theta there is interval of 2pi for +ve n 2n+1 for -ve is it the same for sin also??
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
So the hypotenuse and adjacent both are the same while the altitude is 0. Cos(theta)= adjacent / hypotenuse = 1
Hope this can help: (or this definition from Google: a line that continually approaches a given curve but does not meet it at any finite distance)