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## Trigonometry

### Course: Trigonometry > Unit 2

Lesson 4: Special trigonometric values in the first quadrant# 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?(70 votes)
- 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(130 votes)

- I'm stuck on a problem on the special angles because I don't know how to make a squareroot sign.(34 votes)
- I believe sqrt(x) would be acceptable on khanacademy.(132 votes)

- At5:14, Why would the sqrt of 1/2 be 1/sqrt2 and not sqrt 1/2?(34 votes)
- 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).(101 votes)

- 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.(54 votes)- 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 . :-)(38 votes)

- At7:36, how is the tangent equal to the slope of the hypotenuse? Is it always the case?(17 votes)
- 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).(38 votes)

- Could you also please help with the exercises because they don't look too much related to me.(8 votes)
- At5:15, 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?(15 votes)
- 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.(0 votes)

- 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(12 votes)
- 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

thanks in advance(7 votes)- You are correct Unit 11 lesson 4 is incomplete.

If you need find any other resources you always

a) Google "Special triangles"

b) Use YouTube

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(7 votes)

- can we take pie as 180 in trig functions so that it will easy to term angles in some problems?(5 votes)
- Yes, π radians is equal to 180 degrees. You have to make sure that you understand what you are doing though. If you put in an input in degrees, your answer will be in degrees. If you put in an input in radians, your answer will be in radians.(11 votes)

## Video transcript

Voiceover:So we have depicted
here is a unit circle centered at point A and the
point B lies on the circle. And then we dropped a perpendicular from point B to point D, point D
lies on the positive X-axis. So they form this triangle
ABD, and they tell us at angle BAD, angle bad, it has a
measure of pi over four radians. What I want to do, in this
video, is use our knowledge of trigonometry and use
our knowledge of triangles in order to figure out several things. So the first thing we want
to figure out is what's the radian measure of angle ABD? Actually let's just do that
first and then I'll talk about the other things. that
we need to think about. So I assumed you've paused the video and tried to do this on your own. So let's think about what ABD would be. We know that two of the
angles of this triangle, so if you know two of the
angles of this triangle, you should be able to
figure out the third. Now what might be a little
bit unfamiliar, we're use to saying that the sum
of the interior angles of a triangle add up
to 180 degrees, but now we're thinking in terms of radians. So we could say that the sum of the angles of a triangle add up to,
instead of saying 180 degrees, 180 degrees is the
same thing as pi radians. So this angle plus that angle
are going to add up to pi. So lets just say that
this right over here, lets just say measure
of angle ABD in radians, plus pi over four, plus,
this is a right angle. What would that be in radians? Well a right angle in radians, a 90 degree angle in radians, is pi over two radians. So plus pi over two. When you take the sum of
them, the interior angles of this triangle, they're
going to add up to pi radians. Which is of course the
same thing as 180 degrees. And now we can solve for
the measure of angle ABD. Measure of angle ABD is equal to pi minus pi over two, minus pi over four. I just subtracted these
two from both the sides. So this is going to be
equal to, you can put a common denominator of
four, then this is four pi over four, this is minus two pi over four, and this, of course,
is minus pi over four. So four minus two minus one
is going to get us to one. So this is going to
get us to pi over four. So the measure of angle
ABD is actually the same as the measure of angle
BAD, it is pi over four. So that angle right over
there is pi over four. Now what does that help us with? So if we know that this is pi over four and that is pi over four
radians, and once again we know that this is a unit circle. So we know the length
of segment AB, which is a radius of the circle or is
the radius of the circle is length one, what else do we
know about this triangle? Can we figure out the
lengths of segment AD and the length of segment DB? Well sure, because we have two base angles that have the same measure,
that means that the corresponding sides are also
going to have the same measure. That means that this side is going to be congruent to that side. I can reorient it in a way that might make it a little easier to realize. If we were to flip it over,
maybe not completely fit it over but if we were to make it look like this. So the triangle, we
could make it look like, a little bit more like this. Actually I want to make it
look like a right angle though. So my triangle, let me make
it look like, there you go. So if this is D, this is B, this
is A, this is our right angle. Now this is pi over four radians and this is also pi over four radians. When your two base angles are the same, you know you're dealing
with an isosceles triangle. Because they're not all the
same, it's not equilateral. If all the angles were the
same this would be equilateral. This is an isosceles
non-equilateral triangle. So if your base angles
are the same, then you also know that the corresponding
sides are going to be the same. These two sides are the same,
this is an isosceles triangle. So how does that help us figure out the lengths of the sides? Well if you say that
this side has length X, that means that this side has length X. If this side has length X,
then this side has length X. and now we can use the
Pythagorean Theorem. We can say that this X
squared plus this X squared is equal to the hypotenuse
square is equal to one square. Or we could write that
two X squared is equal to one or that X squared
is equal to one over two or taking the principal
root of both sides, we get X is equal to one
over the square root of two. And a lot of folks don't like
having a radical denominator, they don't like having a rational
number in the denominator. So we could rationalize the denominator by multiplying by the square root of two over the square root of two,
which would be, let's see the numerator will have
the square root of two, and the denominator
we're just going to have square root of two times
square root of two is just two. So we've already been able to figure out several interesting things We've been able to figure
the angle of ABD in radians. We're able to figure out
lengths of segment AD and the length of segment BD. Now what I want to do
is figure out what are the sine, cosine, and tangent
of pi over four radians, given all of the work we have done? So let's first think
about, what is the sine? We'll do this in the color orange. Given all the work we have done, what is the sine of pi over four radians? And I encourage you once
again, pause the video, think about the unit circle definition of trig functions, and
think about what this is. Well the unit circle
definition of trig functions in this angle, this pi over
four radians, is forming an angle with the positive X axis. And where it's terminal ray
intersects the unit circle, the X and Y coordinates of this point are what specify the cosine and sine. So the coordinates of
this point are going to be the cosine of pi over four
radians is the X coordinate and the sine of pi over
four radians is the Y coordinates, sine of pi over four. So what's the Y coordinate going to be if we want pi over four? Well it's going to be this
length, right over here. Which is the same thing as this length, which is the length of X,
which is the square root of two over two. Now what is the cosine of pi over four? Once again I encourage you to pause the video to think about it. What's this X coordinate? The X coordinate, is this
distance right over here, which is once again going to be X, which is square root of two over two. Now, what is the tangent of
pi over four going to be? Well the tangent of pi over
four is sine of pi over four over cosine of pi over four. Now both of these are
the exact same thing. They're both square root of two over two. So you're going to have
square root of two over two, divided by two over two, well
that's just going to give you one. And that also makes
since, because remember the tangent of this angle
is the slope of this line. And we see the slope,
for every X we move in the horizontal direction we move X up. So our change in Y over
change in X is essentially X over X, which is equal to one.