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Algebra 2
Course: Algebra 2 > Unit 11
Lesson 3: The Pythagorean identityProof of the Pythagorean trig identity
CCSS.Math: ,
The Pythagorean identity tells us that no matter what the value of θ is, sin²θ+cos²θ is equal to 1. We can prove this identity using the Pythagorean theorem in the unit circle with x²+y²=1. Created by Sal Khan.
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How is tan theta= cosine theta over sin theta?(10 votes)- Think back algebra. Slope of a line is y/x. Tan represents that slope. Since sin θ is y and cos θ is x. So sin θ/cos θ = y/x = tan θ.(19 votes)
Is there an alternative way to prove the Pythagorean trig identities (there are three of them)?(7 votes)
We can use the definitions of the functions and the following property: x^2 + y^2 = r^2.
sinθ = y/r
cosθ = x/r
sin²θ + cos²θ
= (y/r)² + (x/r)²
= y²/r² + x²/r²
= (x² + y²)/r²
= r²/r²
= 1
We have proven sin²θ + cos²θ = 1. To get to the other identities, all we have to do is divide by sin²θ or cos²θ.
(sin²θ + cos²θ = 1)/cos²θ
tan²θ + 1 = sec²θ
(sin²θ + cos²θ = 1)/sin²θ
1 + cot²θ = csc²θ(25 votes)
Doesn't this proof. only prove the Pythagorean Identity in the unit circle? If the radius was not one, wouldn't this not work.(3 votes)
I had the same question. After a while I figured that it would still work because of how ratios work. Basically to have any other circle you would have to multiply by the same factor:
sin²Θ + cos²Θ = 1
(sin²Θ + cos²Θ)*factor = 1*factor(for different radius)
If you divide each side by the factor, you're back where you started.
I know this answer is super late, but I hope someone else can learn from it...I hope this is correct.(16 votes)
At5:30, why does he put absolute value around the sin and cos?(6 votes)- Good question! Actually, I don't see why Sal would use the absolute value of sin and cos as the square of a negative integer is equal to the square of the absolute value of the same integer.
-a^2 = a^2(7 votes)
What if the radius is not 1?
cos^2(theta) + sin^2(theta) = 4
Do we just divide both sides of the equation by 4?(5 votes)
For the pythagorean trig identity? It works for any radius circle. Try going throught he video again but instead of x^2 + y^2 = 1, pretend that it is x^2 + y^2 = a^2 where we can insert any number for a, so it would be applicable for any circle.
The one twist is that where he has a point's coordinates be cos(theta), sin(theta), it would be a*cos(theta), a*sin(theta) since each point is found by manipulating cos(theta) = x/a and sin(theta) = y/a. let me know if you do not understand why this is.
You would not wind up with cos^2(theta) + sin^2(theta) = 4. Keep in mind x^2 + y^2 = 1 is the same as x^2 + y^2 = 1^2, because int he equation for a circle the number at the end is the radius squared.(3 votes)
Why are the exponents on sine and cosine and not theta?(3 votes)
We do this to remove some ambiguity. If we had sinx^2 instead of sin^2x when we wrote it, you could either regard it as (sin(x))^2 or sin(x^2), which would give you different values. When the exponent is on the trig function, it means to raise the output of the function to the power. However, this solution can also lead to confusion with the inverse trig functions.
We write inverse sine as sin^-1(x). However, raising anything to the -1 is the same as dividing 1 by it. So, we can write sin^-1(x) as 1 / sin(x), which is equal to the cosecant of x. The cosecant is in no way the same thing as the inverse sine, so we have yet more confusion.
Hope this sheds light on the issue.(4 votes)
- 1:50, I don't know if I am misunderstanding or making my own mistake in thought, but shouldn't the cos of theta be the green dotted line and sin of theta be the pink dotted line? Isn't cos of theta the base of the triangle in reference to the triangle the x-axis and the ray make?(5 votes)
This was my big doubt too... but cos is the x value and sin the y value. The pink line connects the point you want to find with the x axis (it's a kind of projection of that point on the line), while the green line connects the point with the y axis... hope this helps!(2 votes)
I know the trig identities, but am having trouble applying them. Can someone help me by explaining the following equation to me please?
(secθ+tanθ)(secθ-tanθ)=1(3 votes)- In general, if you have a product (a+b)(a-b), you can expand it out as a²-b².
So if you rewrite the left-hand-side this way, your equation becomes sec²θ-tan²θ=1. Adding the tangent term to both sides turns this into
tan²θ+1=sec²θ
which is one of the Pythagorean identities.
If you prefer, you can go further by multiplying both sides by cos²θ to get sin²θ+cos²θ=1.(3 votes)
Is there a difference between sin²θ and (sin θ)²?
I cannot seem to understand the difference. It's also a bit confusing to put the exponent there as -1 is used to note the inverse function.(2 votes)- Regarding the confusing use of superscript
-1to indicate an inverse trig function, you can choose to use the "arc" names such asarcsin-- especially in your own notes, or in computer science contexts where the "arc" names predominate.
The Wikipedia article on inverse trig functions uses the "arc" notation, for example: https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Notation
It's still important that you be able to use both notations, but we aren't always "stuck" with the inferior-1notation.(2 votes)
Is sin^2(x) = sin(x)^2 and the same for cos. I am a bit confused?(3 votes)
sin²(x) is the same thing as sin(x)², as is cos²(x) = cos(x)².
This notation was adopted to avoid confusion between sin(x²) and sin(x)². Why?
Often we just write sinθ rather than sin(θ), so now it is easier to see the difference between sinθ² and sin²θ.
When I first saw sin²(x) and sin²x, I preferred to write (sin(x))², but quickly got used to writing less symbols. You will too.(2 votes)
5:30Video transcript
Voiceover: Let's review the unit circle definition of trig functions a little bit. Right over here I've draw a unit circle, and when we say a unit circle we're talking about a circle with radius one. For example this point right over here is the point one comma zero. X is equal to one, Y is zero. This point is the point zero comma one. This is the point negative one comma zero and this is the point zero comma negative one. The radius over here, the distance from the center of the circle which is at the origin to any point in the circle or any point on the circle, I should say, this radius is equal to one. The unit circle definition of trig functions leverages this unit circle, that's why it's called the unit circle definition, and we saw that if we define an angle as the bottom side of the angle being along the positive x-axis and then the other side of that angle, thinking about where it intersects the unit circle. Let's say this is the angle theta. We define sine of theta and cosine of theta, or cosine theta and sine of theta as the x and y coordinates of this point at which this side of the angle, the side that is not the positive x-axis, where that intersects the unit circle. For example this point right over here. We would call this, the x coordinate of this point, so this value right over here, we would call that cosine of theta. The y coordinate of that point, which is this point right over here, we would call that sine of theta. In the previous videos on the unit circle we talked about why this is really just a natural extension of the so cah toa definition. What's useful is it starts to work for negative angles, it even works for 90 degree angles, it works for angles more than 90 degrees, it works for angles less than 90 degrees, so it's really, really, really useful. What I want to do is leverage what we already know about the unit circle definition of trig functions to help prove the Pythagorean identity. It almost falls out of the fact that this point right over here is on a circle, a circle of radius one. What is the equation of a circle with radius one centered at the origin? Well the equation of that is x squared, x squared ... We have other videos where we really prove this using the distance formula, which is really just an application of the Pythagorean theorem. The equation of a unit circle centered at the origin is x squared plus y squared is equal to one, is equal to the radius squared. This distance right over here is equal to one. We've already said that we're defining cosine of theta as the x coordinate of this point and we're defining sine of theta at the y coordinate of this point and this point is sitting on the circle. It has to satisfy this relationship right over here. That means if we're defining cosine of theta to be the x, to be this x value, sine of theta to be the y value and it has to satisfy this relationship that means that cosine of theta squared plus sine squared of theta needs to be equal to one. Or sine square theta plus cosine squared of theta needs to be equal to one. That's just from the point. This is the x, cosine theta is the x coordinate, sine theta is the y coordinate. They have to satisfy this relationship which defines a circle so cosine squared theta plus sine squared theta is one. This is called, as we've seen in other videos, this is called the Pythagorean identity. You say, "Why is that useful?" Well using this, if you know sine of theta you can figure out what cosine of theta is going to be, or vice versa. And if you know one of cosine of theta and then you could -- Say you know cosine theta then you use this to figure out sine of theta, then you can figure out tangent of theta because tangent of theta is just sine over cosine. If you're a little bit confused as to why this is called the Pythagorean identity, well it really just falls out of where the equation of a circle even came from. If we look at this point right over here, we look at this point right over here, which we're saying is the x coordinate is cosine theta and the y coordinate is sine of theta, what is the distance between that point and the origin? Well to think about that we can construct a right triangle. This distance right over here. So that we could deal with any quadrant I'll make it the absolute value of the cosine theta is this distance right over here. And this distance right over here is the absolute value of the sine of theta. I obviously don't have to take the absolute value for this first quadrant here but if I went into the other quadrants and I were to setup a similar right triangle then the absolute value is at play. What do we know from the Pythagorean theorem? This is a right triangle here, the hypotenuse has length one, so we know that this expression squared, the absolute value of cosine of theta squared, plus this expression squared, which is this length, plus the absolute value of the sine of theta squared needs to be equal to the length of the hypotenuse squared, which is the same thing which is going to be equal to one squared. Or we could say, this is the same thing. If you're going to square something the sign, if negative it's going to be negative times a negative so it's just going to be positive so this is going to be the same thing as saying that the cosine squared theta plus sine squared theta is equal to one. This is why it's called the Pythagorean identity. Actually that's even where the equation of a circle comes from, it comes straight out of the Pythagorean theorem where your hypotenuse has length one.