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### 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.

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• How is tan theta= cosine theta over sin theta?
• 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.
• 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?
• 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
• wait, I understand that the hypotenuse and the x and y are 1, but how does 1+1=1?
• The unit circle does not go through (1,1). At a 45 degree angle where sin and cos are equal, the point is (√2/2,√2/2).
• at , when khan say they proved that x^2+y^2=1 equals the equastion for a unit circle, can someone explain that or show me a vid that can help me understand.

Thanks!
• If you have a point (x, y), the distance from the origin to (x, y) is given by √(x²+y²).

The unit circle is the set of points where this distance is 1, where √(x²+y²)=1. Squaring both sides of the equation, this is equivalent to x²+y²=1.
• 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.
• Does this only work for 90-degree triangles?
• No, whatever the value of θ is, sin^2(θ) + cos^2(θ) = 1
• 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.