Trigonometric Identities at a Glance

Proving the Pythagorean Identity

Remember Pythagoras? You know...old...Greek...dead? The guy who loved triangles? The gentleman that came up with the Pythagorean Theorem? Yup, that's the one. Well, he passed away looking a little like this:

Pythagoras (our imaginary version at least) held onto something called the Pythagorean identity way after his death, because it was right. And it's still as right today as it was back then, unlike his wardrobe. The Pythagorean identity states that:

sin² ɵ + cos² ɵ = 1

So what's the big deal? The big deal is that the Pythagorean identity relates sine and cosine in a very simple way. That will definitely come in handy: imagine trying to find an angle's cosine without any of the triangle's measurements except for sine. The Pythagorean identity can handle that and so much more. But where does the Pythagorean identity even come from? We want to see a proof.

Let's start with a picture of a well-labeled right triangle (the pointy kind), making sure, of course, the triangle sits in the unit circle.

Well, we know a few things about this triangle right away. We know the sum of the square of the legs x and y equals the square of the hypotenuse r (thank you, Pythagoras).

x² + y² = r²

We also know the value of the sine and cosine of θ in terms of x, y, and r (thank you, trig ratios). 

sin ɵ = ^y/_r

cos ɵ = ^x⁄_r

Oh, and let's not forget that since the triangle is in the unit circle, the hypotenuse equals 1. 

r = 1

Now, we can take this info to prove that the Pythagorean identity is true.

Let's start with this equation:

x² + y² = r²

Now, let's substitute x, y, and r for the values we know them to be. We know r equals 1, so we can rewrite it:

x² + y² = 1²

Go back to those sine and cosine equations real quick. Let's solve for x and y in those two equations.

sin ɵ = ^y/_r
y = r sin ɵ

cos ɵ = ^x⁄_r
x = r cos ɵ

Plug 'em into the basic Pythagorean Theorem equation.

x² + y² = 1²

(r cos ɵ)² + (r sin ɵ)² = 1²

Hmmm...that looks vaguely familiar. But it's not there yet. Let's simplify.

r²(cos ɵ)² + r²(sin ɵ)² = 1²

r²(cos² ɵ + sin² ɵ) = 1²

But hey, we already decided that r = 1.

1²(cos² ɵ + sin² ɵ) = 1²

cos² ɵ + sin² ɵ = 1²

Since 1² equals 1, we've finally made it:

cos² ɵ + sin² ɵ = 1

High fives all around.

Sample Problem

If sin ɵ = 0.65, what's the cosine of ɵ?

Looks like a job for the Pythagorean identity. First we'll jot down our brand-new, piping-hot formula:

sin² ɵ + cos² ɵ = 1

Now just plug in what we know.

(0.65)² + cos² ɵ = 1

0.4225 + cos² ɵ = 1

So far, so good. Let's solve for the cosine.

cos² ɵ = 1 – 0.4225

cos² ɵ = 0.5775

That's the cosine squared though, so we'll need to grab the square root of both sides to finish up.

cos ɵ ≈ 0.760

Sample Problem

Given with ɵ in Quadrant IV, find tan ɵ.

First, sketch your problem. We know cosine is the adjacent side over the hypotenuse, so we can label two sides in our triangle.

x = 9
y = ?
r = 41

This time around, we can find y using the Pythagorean Theorem.

But remember that we're in Quadrant IV, and y-values are negative there.

y = -40

Plugging into our tangent trig function gives us:

Sweet.

Sample Problem

Given ɵ in Quadrant II and , find sin ɵ. 

Draw a sketch like this to help you out.

Now we can clearly see that x = -28, y = 45, and r = ? (otherwise there would be no problem).

First, let's find r.

So we have the required ingredients to use our sine trig function and get our answer:

Sample Problem

Given in Quadrant III, find cos ɵ.

Make a sketch like this:

x = ?
y = -91
r = 109

Find x:

But since we're in Quadrant III, slap a negative sign on there.

x = -60

Plugging into our cosine trig identity gets us: