The trig functions & right triangle trig ratios
Sal shows how, for acute angles, the two different definitions of the trigonometric values (SOH CAH TOA and the unit circle definition) result in the same values. Created by Sal Khan.
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- Hard to pay attention to, all I see is letters and those don't add or divide. they are not numbers. letters everywhere and confused(35 votes)
- Think of them as numbers with no identity, they are called variables.(4 votes)
- Still kinda confused on the fractions on the unit circle, hard to remember.(15 votes)
- is to illustrate the fact their either the unit circle or a plain triangle can be used to find the same trig ratios(6 votes)
- The video doesn't really help with the practice exercises, i've seen similar complaints below that are from a year ago and the issue has not been fixed since then... i'm jumping to the next section for now but please fix the issue, so we can come back and finish this section.
I've got two exercises out of five right in a row and as soon as i think to have got it, something blows up, it's so frustrating...(14 votes)
- I found that watching the videos on Symmetry and periodicity of trig functions https://www.khanacademy.org/math/trigonometry/less-basic-trigonometry/trig-symmetry-periodicity/v/trigonometry-unit-circle-symmetry really helped with the Unit Circle Trigonometry practice.(11 votes)
- In the unit circle, can we imagine the hypotenuse in any rt angle triangle is a straight line on a cartesian plane and so it's equation is y=mx+c, and, since c = zero, m=y/x - the equation for tan(theta). So tan(theta) is the gradient of the hypotenuse?(8 votes)
- Yes, your reasoning is perfectly right. Tangent is defined by the Y ordenade of the point in which the line defined by the angle crosses the x=1 line.(2 votes)
- What does Sal mean by the Unit Circle is an extension of SOH-CAH-TOA? I know what they mean, but how is it an extension?(3 votes)
- In SOH, CAH, TOA, definition of trig functions, we defined the sin,cos etc .
as the ratio of the sides of a triangle. Also, we were only able to find the value of trig functions of angles upto 90 degrees.
But in unit circle definition, the trigonometric functions sine and cosine are defined in terms of the coordinates of points lying on the unit circle x^2 + y^2=1. And unlike the previous definition(SOH, CA...), we are able to find sin,cos.... of all possible angles.
That's how the Unit Circle definition is an extension of SOH,CA.... definition.(8 votes)
- Not much explanation of what the ratios are or how they are used. Just a lot of quick talk. The exercise questions do not follow the lesson and have that bland text-book feel to them when you use hints. Khan Academy has definitely moved away from the simple approach that it was founded on to a more public school diploma chugging factory approach...(6 votes)
- How can I solve for a trigonometric function without using my calculator? What does mathematicians do to solve a sine or inverse sine function in middle ages?(4 votes)
- According to Wikipedia, "Before the existence of pocket calculators, trigonometric tables were essential for navigation, science and engineering."(4 votes)
- So suppose ive been given that sin(theta) is -4/5 and that cos(theta) > 0, how would I find cos(theta) and tan(theta)(2 votes)
- You could use the Pythagorean identity: sin²(θ) + cos²(θ) = 1
First you rearrange that to give an formula for cos(θ) in terms of sin(θ) - which you know.
Then, once you have cos(θ), you can get tan(θ) = sin(θ)/cos(θ)(4 votes)
- Still kinda confused on the fractions on the unit circle, hard to remember.(2 votes)
- Which fractions are you talking aobut, the trig fractions as shown by SOH CAH TOA, fractions created by points at various angles such as 30, 60, 90, etc. or the fractions in radians related to these degree angles? There are all sorts of fractions related to the unit circle. I also notice that this is not even your own question, it is directly copied from the second most popular post, come up with origin question because directly copying and acting like it is your own is plagiarism.(4 votes)
- How are the values of the trig values derived? How could you get what sin or tan or cos of an angle was without a calculator?(2 votes)
- There is no practical way to do it without a calculator
Before calculators, people used big "trig tables" printed on paper(3 votes)
Voiceover:On the right-hand side we have a bunch of expressions that are just ratios of different information given in these two diagrams. Then over here on the left we have the sine taken of angle MKJ, cosine of angle MKJ, and tangent of angle MKJ. Angle MKJ is this angle right over here same thing as theta, so these two angles. These two angles have the same measure. We see that right over there. What we want to do is figure out which of these expressions are equivalent to which of these expressions right over here. I encourage you to pause the video and try to work this through on your own. Assuming you've had a go at it, let's try to work this out. When you look at this diagram, it looks like the intention here on the left is this evokes the unit circle definition of trig functions because this is a unit circle right over here, and this evokes kind of the soh cah toa definition because we're just kind of in a plain, vanilla right triangle. Just to remind ourselves, let's just remind ourselves of soh cah toa because I have a feeling it might be useful. Sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse, and tangent is opposite over adjacent. We can refer to this and we can also remind ourselves of the unit circle definition of trig functions that the cosine of an angle is the X coordinate and that the sine of where this ray intersects the unit circle, and the sine of this angle is going to be the Y coordinate. What we'll see through this video is that they actually, the unit circle definition, is just an extension of soh cah toa. Let's look first at X over one. We have X, X is the X coordinate. That's also the length of this side right over here, relative to this angle, theta. That is the adjacent side. So X is equal to the adjacent side. What is one? Well, this is a unit circle. One is the length of the radius which for this right triangle is also the hypotenuse. If we apply the soh cah toa definition, X over one is adjacent over hypotenuse, adjacent over hypotenuse, adjacent over hypotenuse, that's cosine. That's going to be this is equal to cosine of theta, but theta is the same thing as angle MKJ. They have the same measure so cosine of angle MKJ is equal to cosine of theta which is equal to X over one. Now let's move over to Y over one. Well, Y is going to be the length of this side right over here. Y is going to be, let me do this in the blue. Y is going to be this length relative to angle theta. That is the opposite side. That is the opposite side. Now which trig function is opposite over hypotenuse? Opposite over hypotenuse? That's sine of theta. Sine of theta. So sine of angle MKJ is the same thing as sine of theta. We see that they have the same measure, and now we see that's the same thing as Y over one. Now for both of these I used the soh cah toa definition, but we could have also used the unit circle definition. X over one, that's the same thing. That's the same thing as X, and the unit circle definition says the X coordinate of where this, I guess you could say, the terminal side of this angle, this ray right over here, intersects the unit circle. That by definition, by the unit circle definition is the cosine of this angle. X is equal to the cosine of this angle, and the unit circle definition, the Y coordinate is equal to the sine of this angle. We could have written this as instead of X, Y, we could have written this as cosine of theta, sine theta just like that, but let's keep going. Now we have X over Y. We have adjacent over, we have adjacent over opposite. So this is equal to adjacent over opposite. Tangent is opposite over adjacent, not adjacent over opposite. This is the reciprocal of tangent. This right over here, if we had to, this is equal to one over tangent of theta. We later learn about cotangent and all of that which is essentially this, but it's not one of our choices. So we can rule this one out. Then we have Y over X. Well, this is looking good. This is Y is opposite. Opposite. X is adjacent relative to angle theta. Adjacent. So this is the tangent of theta. This is equal to tangent of theta. Tangent of angle MKJ is the same thing as tangent of theta which is equal, which is equal to Y over X. Now let's look at J over K, so J over K. Now we're moving over to this triangle, J over K. Relative to this angle because this is the angle that we care about, J is the length of the adjacent side, and K is the length of the opposite side, of the opposite side. This is adjacent over opposite. This is equal to adjacent over opposite. Tangent is opposite over adjacent not adjacent over opposite. So once again this is the reciprocal of the tangent function not one of the choices right over here so we can rule that one out. Now K over J. Well, now this is opposite over adjacent. Opposite over adjacent. That is equal to tangent of theta. This is equal to tangent of theta, or tangent of angle MKJ. This is equal to K over J. Now we have M over J, M over J. Hypotenuse over adjacent side. This of course is equal to the hypotenuse. Hypotenuse over adjacent. Well, if it was adjacent over hypotenuse, we'd be dealing with cosine, but this is the reciprocal of that. This is actually one over the cosine of theta not one of our choices, not one of our choices here so I'll just rule that one out over there. Then we have it's reciprocal, J over M. That's adjacent over hypotenuse. Adjacent over hypotenuse is cosine. This is equal to cosine of theta, or cosine of angle MKJ so we could write it down. This is equivalent to J over M. Then one last one, K over M. Well, that's opposite over hypotenuse, opposite over hypotenuse. That's going to be sine of theta. This right over here is equal to sine of theta which is the same thing as sine of angle MKJ which is the same thing as all of these expressions. This is equal to K over M, and we are done.