- [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 three-fourths,
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.