Main content
Precalculus
Cosine, sine and tangent of π/6 and π/3
With the unit circle and the Pythagorean theorem, we can find the exact sine, cosine, and tangent of the angles π/6 and π/3. Created by Sal Khan.
Want to join the conversation?
 How does Sal determine that the point of which π/3 intersects the actual unit circle in in? I still don’t understand that especially the part of which he says that it is (cos π/3,sin π/3)? Please let me know, thank you. 4:10(9 votes)
 So π/3 is 60 degrees (π/3*180/π) which is how he estimates about where π/3 is. He then uses trig functions to get the points. By drawing a right triangle, the hypotenuse is 1 (radius of unit circle), the adjacent part along the x axis is defined by the function cos(π/3) = adj/hyp, but since the hyp=1, you get adj = cos(π/3) and the opposite part of the triangle would be sin(π/3) = opp/hyp, so the opp =sin(π/3). So staying in the first quadrant, the point on the unit circle defined by 0<x<π/2 (or 0<x<90) would always be (cos(x),sin(x)).(10 votes)

 just… get better, lol jk try here we go. embracing the challenge of trigonometry with a positive and determined mindset. Visualize concepts through diagrams and realworld applications to make the subject more engaging and relevant. Set achievable goals and celebrate each milestone, no matter how small. Seek support when needed, and remember that patience and persistence are key to mastering trigonometry. Stay positive, avoiding selfdoubt, and view difficulties as opportunities to learn and grow. Draw inspiration from the achievements of mathematicians and scientists who have utilized trigonometry to solve fascinating problems throughout history. With dedication and motivation, you can unravel the beauty and practicality of trigonometry, and its applications will inspire your mathematical journey.
I really hope this help mate :)) stay strong!(6 votes)
 just… get better, lol jk try here we go. embracing the challenge of trigonometry with a positive and determined mindset. Visualize concepts through diagrams and realworld applications to make the subject more engaging and relevant. Set achievable goals and celebrate each milestone, no matter how small. Seek support when needed, and remember that patience and persistence are key to mastering trigonometry. Stay positive, avoiding selfdoubt, and view difficulties as opportunities to learn and grow. Draw inspiration from the achievements of mathematicians and scientists who have utilized trigonometry to solve fascinating problems throughout history. With dedication and motivation, you can unravel the beauty and practicality of trigonometry, and its applications will inspire your mathematical journey.
 Hello
I just completed composite functions in Precalculus and the videos were really informative. However, the next section, which is this (trigonometry), is a little confusing. I don’t understand the video above that I just watched, am I lacking some sort of foundational math that someone can refer me to before coming on to this section? I took elementary math in high school and not additional math which covers precalculus to calculus and that’s why I am learning precalculus now before moving on to calculus but I don’t recall learning this before.
Thank you :)(5 votes) So you are still in the precalculus section but in the trignometry unit.
Instead go to trigonometry section and start at the first unit. There you will learn trigonometry from the start.(6 votes)
 So you are still in the precalculus section but in the trignometry unit.
 Does that mean if hypotenues = A then adjecent side is A.cos(theta) and opposite side is A.sin(theta)?(3 votes)
 Yes, since sin(<)=opp/hyp, you could multiply by hyp to get opp=hyp sin(<), same for cos(<).(6 votes)
 I think this video also needs to appear in the Algebra 2 curriculum. Right now Unit 11 Lesson 4 of Algebra 2 only has the video “Trig values of π/4” and the exercise “Trig values of special angles.”(5 votes)
 Many parts of math overlap. Nothing wrong about that.(2 votes)
 I recently saw a question on someone else’s test that I wasn’t sure about.
csc 158° 10'
I assume that the problem was saying 10 * csc(158), where the tick mark mean distance or perhaps feet? The only question was to calculate the answer.
I am not aware of the notation with the little tickmark as shown.
Thanks for your time!(4 votes) A degree of arc is subdivided into 60 ‘minutes of arc’, or just ‘minutes’. An arcminute is further divided into 60 arcseconds. So there are 60^2=3600 arcseconds in a degree.
We denote an arcminute with a ‘, and an arcsecond with a “.
So 158º 10′ is 158 degrees, 10 minutes, or 158 and onesixth degrees (since 10/60=1/6).(2 votes)
 A degree of arc is subdivided into 60 ‘minutes of arc’, or just ‘minutes’. An arcminute is further divided into 60 arcseconds. So there are 60^2=3600 arcseconds in a degree.
 How do you find the trig ratios(sine, cosine and tangent) for other angles like pi/5, pi/10 or any other arbitrary angle.(2 votes)
 The two angles you mention, 𝜋∕5 = 36° and 𝜋∕10 = 18°, we can actually find exact expressions for that aren’t too complicated.
We start with a 727236 triangle.
We let the base of this isosceles triangle have length 1, and the legs have length 𝑥.
Then we draw the bisector to one of the 72° angles.
This splits the triangle into two new isosceles triangles, one of which is a 727236 triangle.
We can figure out that this triangle has a base of length 𝑥 − 1, while the legs are of length 1.
Since this triangle is similar to the original triangle, we can set up the equation
𝑥 − 1 = 1∕𝑥.
Multiplying both sides by 𝑥 and then subtracting 1 from both sides, we get
𝑥² − 𝑥 − 1 = 0,
which has the positive solution
𝑥 = (1 + √5)∕2.
Now we go back to the original triangle.
Because it is isosceles, we know that its height bisects both the 36° angle and the base, thus splitting the triangle into two congruent right triangles –
either of which having one angle being 18° with the opposite side having length 1∕2
and a hypotenuse of length 𝑥 = (1 + √5)∕2.
Thus, sin(18°) = (1∕2)∕((1 + √5)∕2),
which simplifies to (√5 − 1)∕4.
Because 18° lies in the first quadrant, we know that cos(18°) is positive:
cos(18°) = +√(1 − sin²(18°)),
which we calculate to √(10 + 2√5)∕4.
To get a somewhat neat expression for tan(18°) we begin by finding
tan²(18°) = sin²(18°)∕cos²(18°) = (1 − cos²(18°))∕cos²(18°) = 1∕cos²(18°) − 1
= 16∕(10 + 2√5) − 1,
which can be simplified to (25 − 10√5)∕5².
Again, 18° lies in the first quadrant, so tan(18°) is positive, and we get
tan(18°) = √(25 − 10√5)∕5.
– – –
To find sin(36°) we can use the double angle formula
sin(2𝑥) = 2 sin(𝑥) cos(𝑥)
⇒ sin(36°) = 2 sin(18°) cos(18°) = 2(√5 − 1)∕4⋅√(10 + 2√5)∕4
= (√5 − 1)√(10 + 2√5)∕8.
Again, squaring both sides allows us to get a neater result:
sin²(36°) = (6 − 2√5)(10 + 2√5)∕64,
simplifying to (10 − 2√5)∕16.
36° lies in the first quadrant, so
sin(36°) = √(10 − 2√5)∕4.
cos(36°) = √(1 − sin²(36°)),
which we calculate to √(6 + 2√5)∕4.
Finally,
tan²(36°) = 1∕cos²(36°) − 1 = 16∕(6 + 2√5) − 1,
which simplifies to 5 − 2√5
⇒ tan(36°) = √(5 − 2√5).(6 votes)
 The two angles you mention, 𝜋∕5 = 36° and 𝜋∕10 = 18°, we can actually find exact expressions for that aren’t too complicated.
 Why is tangent = Sine over cosine?(1 vote)
 sine = opposite/hypotenuse
cosine = adjacent/hypotenuse
sine∕cosine = (opposite/hypotenuse)/(adjacent/hypotenuse)
= opposite∕adjacent = tangent(7 votes)
 sine = opposite/hypotenuse
 how did he get 3/4 and then b sqaure root of 3/2?(3 votes)
 what you are referring to is the special triangles. In this case the equilateral triangle with side 1. Using right angle trigonometry and geometry you can derive values of trig functions at pi/3 and pi/6.
It is bit hard for me explain, but you have question go over Sal’s video (this video) and feel free to ask question regarding working out/steps.(3 votes)
 what you are referring to is the special triangles. In this case the equilateral triangle with side 1. Using right angle trigonometry and geometry you can derive values of trig functions at pi/3 and pi/6.
 : How does Sal get b is the sqrt of 3 over 2? He doesn’t seem to show his work there… 3:25(2 votes)
 He starts from b^2 = 3/4. The opposite of squaring is square rooting (sort of, but Sal mentions we are taking the primary positive – root only). So taking square root of everything, you get √b^2 = √3/√4. On the left, the square and root cancel, on the right √4=2, so you end up with b = √3/2.(4 votes)
Video transcript
– [Instructor] In this video, we’re going to figure out what the sine, cosine and tangent of two very important angles are. Angles that you will see a lot in your trigonometric and just in general in your regular life. So these are the angles, pi over 3 radians and pi over 6 radians. And sometimes it’s useful to visualize them as degrees. pi over 3, you might remember pi radians is 180 degrees, so you divide that by three, this is equivalent to 60 degrees. And once again, 180 degrees, which is the same thing as pi radians divided by six is the same thing as 30 degrees. Now, I’m going to do it using the unit circle definition of trig functions. But to help us there, I’m going to give us a little bit of a reminder of what some of you might be familiar with as 30, 60, 90 triangles, which I guess we could also call pi over six, pi over three, pi over two triangles. And so let me just draw one because this is going to be really helpful in establishing these trig functions using the unit circle definition. So let me draw a triangle here, it’s hand drawn, so it’s not as neat as it could be. So this right over here is a right angle, and let’s say that this one is pi over three radians which is the same thing as 60 degrees, and this one over here is pi over six radians which is the same thing as 30 degrees. Now, let’s also say that the longest side, the hypotenuse here has length one. Now, to help us think about what the other two sides are, what I’m going to do is flip this triangle over this side right over here, and essentially construct a mirror image. So because this right over here is a mirror image, we immediately know a few things. We know that this length right over here is going to be congruent to this length over here. And let me actually finish drawing the entire triangle, it’s going to look something like this. And since once again, it’s a reflection, this length over here is going to have length one, this is going to be pi over six radians, this is going to be pi over three radians. So what else do we know about this larger triangle now? Well, we know it’s an equilateral triangle. All the angles, pi over three radians, pi over three radians and if you add two pi over sixes together, you’re going to get pi over three as well, so it’s a 60 degree, 60 degree, 60 degree triangle. And so all the sides are going to have the same length, so it’s going to be one, one and one. And if these two sides are congruent of the smaller triangles, of the smaller right triangles, well then this right over here must be one half, and then this right over here must be one half as well. Now, that’s going to be useful for figuring out what this length right over here is going to be. Because we have two right triangles, we could use either one, but if we just use this bottom right triangle here the Pythagorean theorem tells us that one half squared, let’s call this B, so plus B squared, I’m just pattern matching, A squared plus B squared is equal to C squared where C is the length of the hypotenuse, is equal to one squared. And so we get that one fourth plus B squared is equal to one or subtracting one fourth from both sides. B squared is equal to threefourths, and then taking the principle root of both sides, we get B is equal to the square root of three over two. So just like that, we have figured out what all the lengths of this 30, 60, 90 triangle are. So B here is equal to square root of three over two. Now, I said this would be useful as we go into the unit circle definitions of sine, cosine and tangent. And we’re about to see why. So here I have two different unit circles, I’m going to use one for each of these angles. So first, let’s think about pi over three radians. And so pi over three, would look something like this, this is pi over three radians. And the cosine and sine can be determined by the X and Y coordinates of this point where this radius intersects the actual unit circle. The coordinates here are going to be cosine of pi over three radians and sine of pi over three radians. Or another way to think about it is, I can set up a 30, 60, 90 triangle here, so I’m going to drop a perpendicular. This would be 90 degrees or pi over two radians. And then this angle over here, this is 60, this is 90, this is going to be 30, or another way of thinking about it, it’s going to be pi over six radians. It’s going to be just like one of these triangles here. And so the X coordinate, which is going to be the same thing as the cosine of pi over three, is going to be the length of this side, right over here. Well, what’s that going to be? Well, when your hypotenuse is one, we know that the shorter side, the side opposite the pi over six radians, is one half. So just like that, we have been able to establish that cosine of pi over three radians is equal to one half. This right over here is one half, that is the X coordinate where this radius intersects the units circle. Now, what about the Y coordinate? What is sine of pi over three going to be? Well, the Y coordinate is the same thing as the length of this side, and once again, it goes back to being this triangle. If this is one, this is one half, this is one, this is one half, this other side is going to be square root of three over two. So sine of pi over three is going to be square root of three over two, so let me write that down. Sine of pi over three is equal to square root of three over two. And these are good ones to know. I never say really memorize things, it’s always good to know how to derive things in case you forget. But if you have to memorize them I would highly recommend memorizing these, and then of course from these we can figure out the tangent. The tangent is just going to be the sine over the cosine, so let me write it down here. The tangent of pi over three is going to be the sine, which is square root of three over two, over the cosine which is one half, got a little squanchy down there, and so this is just going to be square root of three over two times two is just going to be square root of three. So now let’s just use that same logic for pi over six. And in fact, I encourage you to pause this video and see if you can do that on your own. All right, now let’s draw a radius that forms a pi over six radian angle with a positive X axis, might look like that. So if that’s going to be pi over six radians, you might imagine it’s interesting to drop a perpendicular here and see what type of triangle we’ve constructed. So this has length one, this is pi over six radians, this is a right angle. So this again, is going to follow the same pattern. This will be pi over three radians. And so the sides are exactly the exact same as this top blue triangle here. So we know that this length over here is going to be one half. We know that this length over here is going to be square root of three over two. And that’s useful because that tells us the coordinates here. The coordinates here, the X coordinate of this point where the radius intersects the unit circle is square root of three over two, and then the Y coordinate is one half. And that immediately tells us the cosine and the sine of pi over six, let’s just write it down. So this tells us that cosine of pi over six is equal to square root of three over two. And sine of pi over six is equal to one half. Notice, we actually just swap these two things around because now the angle that we’re taking the sine or cosine of, is a different angle on a 30, 60, 90 triangle, but we’re essentially utilizing the same side measure, just one way to think about it. And then what’s the tangent going to be? I’ll write it down here. The tangent of pi over six is going to be the sine over the cosine square root of three over two, and so that’s going to be equal to one half times two over the square root of three, which is equal to one over the square root of three. Now some people sometimes don’t like radicals in the denominator and so you can multiply the numerator and the denominator by square root of three if you like to get something like this, you multiply the numerator and denominator by square root of three you get square root of three over three, which is another way of writing tangent of pi over six. But either way, we’re done, it’s very useful to know the cosine, sine and tangent of both pi over three and pi over six. And now you also know how to derive it.