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### Course: Algebra (all content)>Unit 14

Lesson 5: Trigonometric values of special angles

# Trig values of π/4

Sal finds the trigonometric values of π/4 using the unit-circle definition. Created by Sal Khan.

## Want to join the conversation?

• I dont really get the questions in the Trig Values of Special angles activity, cound anyone explain?
• It actually requires more knowledge than Sal has taught us to this point. It's very strange how they've not taught us this in Algebra and say not to use a calculator. I found this link which is very useful https://www.mathsisfun.com/geometry/unit-circle.html
• I'm stuck on a problem on the special angles because I don't know how to make a squareroot sign.
• I believe sqrt(x) would be acceptable on khanacademy.
• At , Why would the sqrt of 1/2 be 1/sqrt2 and not sqrt 1/2?
• This is because the sqrt(1/2) is the same thing as saying sqrt(1)/sqrt(2) and since the square root of 1 is just 1 we can simplify the numerator and rewrite it as 1/sqrt(2).
• Can anyone show how we know that sin(pi/6)=1/2 ? Or from the sin=1/2, how do we get the angle pi/6 ? His explanation for the angle pi/4 makes perfect sense, but do we just have to accept that sin(pi/6)=1/2 and cos(pi/6)=sqrt3/2 ?

It would be nice if someone would at least say: "It was found using extremely difficult and time consuming math that is too much trouble to explain." and then maybe point towards a resource to check it out.

Update:

I figured it out! What nobody here has been saying, because it must have seemed so obvious, is that the angle pi/6 is half of an equilateral triangle! All three angles are 60 degrees (pi/3). Cut it into two right triangles and you get an angle of 30 degrees (pi/6). That also means that the opposite side is going to be exactly half of the hypotenuse. In a unit circle that means that sin=1/2. From there we can work out cos=sqrt3/2

I might be dumb for not seeing the obvious, but there really ought to be a video for this.
• You're not dumb! You figured out an important principle in trigonometry involving the square root of an irrational number. You're also persistent in that you didn't just sigh and quit, you worked at it until a solution became clear. Good for you, Amos . :-)
• At , how is the tangent equal to the slope of the hypotenuse? Is it always the case?
• If you look at the soh cah toa definition of the tan function, we see that the tan is the opposite divided by the adjacent. This is the exact definition of a slope: the amount of rise (opposite) for distance traveled (adjacent).
• Could you also please help with the exercises because they don't look too much related to me.
• for all those stuck in algebra 2 of khan academy and are confused with the next exercise (Trig values of special angles) you need to go watch https://www.khanacademy.org/math/trigonometry/unit-circle-trig-func/xfefa5515:special-trigonometric-values-in-the-first-quadrant/v/cosine-sine-and-tangent-of-6-and-3
• At , Sal finds the side lengths of the isosceles triangle, and that is easy to understand. If the two side lengths are different, how would you find the lengths?
• It's still SOH CAH TOA, but you might have trouble getting your calculator to spit out the right answer unless it's set in radians. Eg if your calculator is set in degrees then Cos(pi/3)=0.86 but if set to radians it'll be 0.5 which actually works.
• what does this video have anything to do with the practice exercises
ill greatly appreciate any help
i dont understand how to do the exercises
• You are correct Unit 11 lesson 4 is incomplete.

If you need find any other resources you always

c) Use a textbook
d) Search through the khan academy Q/A

I will mention this in feedback to khan academy

Edit: It seems like the videos on the trigonometry section are complete but not algebra 2