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## Algebra 2

### Course: Algebra 2>Unit 11

Lesson 3: The Pythagorean identity

# Proof of the Pythagorean trig identity

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.

## Want to join the conversation?

• How is tan theta= cosine theta over sin theta? • Is there an alternative way to prove the Pythagorean trig identities (there are three of them)? •  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
• Doesn't this proof. only prove the Pythagorean Identity in the unit circle? If the radius was not one, wouldn't this not work. • 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.
• At , why does he put absolute value around the sin and cos? • 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? • 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.
• Why are the exponents on sine and cosine and not theta? • 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.
• , 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? • 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 • 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.
• 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.  