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## Precalculus

### Course: Precalculus>Unit 2

Lesson 2: Trigonometric identities on the unit circle

# Sine & cosine identities: periodicity

Sal finds trigonometric identities for sine and cosine by considering angle rotations on the unit circle. Created by Sal Khan.

## Want to join the conversation?

• What are some other relationships besides the one that Sal came up with at the end of the video?
• Here are some trigonometric relationships that can be found by playing around with the unit circle:
sin(pi-θ)=sin(θ)
cos(pi-θ)= -cos(θ)
sin(θ+pi)= -sin(θ)
cos(θ+pi)= -cos(θ)
cos(θ+pi/2)= -sin(θ)
sin(-θ)= -sin(θ)
cos(-θ)=cos(θ)

• Is sin always related to the y axis? I'm confused because if we look at angle theta + pi/2, why is it that sin doesn't have the opposite/hypotenuse definition? If sin is opposite side/hypotenuse wouldn't the sin of theta + pi/2= sin of theta? The opposite side of the "triangles" is the same isn't it?
• Yes, sine is directly related to the y axis. When an angle intersects the unit circle, the sin is equal to the y value of the point at which it intersects.
Sine (theta+pi/2) is equal to cosine.
• Is it me, or did Sal mistake his notation in the final part of the video in concluding that cos(theta) = sin(theta+pi/2)? Because that is not a right triangle, and would need to be analyzing its complimentary angle (90-theta) since 90-theta+theta+pi/2=180 degrees. Since sin(90-theta) = cos(theta), and cos(90-theta) = sin(theta) this seems to be the right way... anyone else notice this or am I mistaken?
• Sine, cosine and the other functions are not just defined for right angles, though the simple definitions you start with for these functions only work for the acute angles of right triangles.

But, yes, cos x = sin(x + ½π)
It is also true that cos x = sin(½π - x)
Thus, it is true that
sin(½π - x) = sin(x + ½π)

It is also true that cos(½π-x) = sin(x)
and that cos(x- ½π) = sin(x)
So, cos(½π-x) = cos(x- ½π)

Sine and cosine are both periodic functions that are identical except for being shifted ½π radians out of phase. Thus, there are a number of ways you can shift them around to be in phase and therefore equal.

Note: Do not mix degrees and radians. Once you can use radians, you should just drop degrees altogether because in math that isn't too far down the road, degrees just won't work, you have to use radians.
• At Sal says that the length of the magenta line is Sine theta plus Pi. How did he come to that?

Wouldn't the length of that line be equal to Cos theta since we are rotating the triangle?
• Yes, you're saying the same thing Sal said - that they're equal.

By using the unit circle definition of sin(a) and cos(a) ( = y/r and x/r at angle a);
sin(t+pi/2) = cost
• does cos(θ+π/2) = sin (θ) ?
• Not quite. Remember the direction in which you want the function to move. Since you want to move the cosine function forward, you have to subtract the pi/2 from the theta value.

So, sin(theta) = cos(theta - pi/2).

What you have there will be -sin(theta)
• Sal says that the unit circle definition is an extension of soh-cah-toa, but wouldn't it be the other way around? Isn't soh-cah-toa an extension of the unit circle definition?
(1 vote)
• I guess you could think of it both ways. Both definitions are derivable from the other definition. For example, you can say:
Sin(x) = Opposite/Hypotenuse
(In a unit circle, Hypotenuse=1, so)

> Sin(x) = Opposite/1
> Sin(x) = Opposite
In this case, the length of the side opposite the central angle = the y length of the triangle.
>> that means that the y coordinate (Where the Hypotenuse intersects the circle) = Sin(x)
>>>the same logic applies for cos(x)

The reason some people say that the unit circle definition is an extension of Soh-Cah-Toa is mostly because Soh-Cah-Toa (I believe) was defined first, and the Unit circle definition is applicable to more scenarios(Any triangle versus only right triangles). The Unit circle definition EXPANDS upon the Soh-Cah-Toa definition. It takes you more places.
• In the video, Sal proved that:

`sin(π/2 + θ) = cos θ`

But, in my textbook, the identity given goes like this;

`sin(π/2 - θ) = cos θ`

So does that mean that `sin(π/2 - θ) = sin(π/2 + θ)` or is there a mistake somewhere?
• Yes, it does mean that.
sin(π/2 - θ) = sin(π/2 + θ) can be interpreted as saying that the sine function has mirror symmetry about the line x = π/2, which I think you'll agree is the case.
• I discovered that sin Θ = cos (Θ+pi/2)........
Along with that I observed that sin (Θ-pi/2) =( sin(Θ+pi/2) / cos(Θ+pi/2) )
Are my observations correct?
• Not true. Try pi/4. Sin pi/4 = 1/sqrt(2) but cos (pi/4 + pi/2)=-1/sqrt(2). You are off by a phase shift of pi (180 degrees out of phase.)

Keep experimenting, but look at the signs of the functions on the unit circle. they are important.