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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:

sin^{2} ɵ + cos^{2 }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).

*x*^{2} + *y*^{2} = *r*^{2}

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:

*x*^{2} + *y*^{2} = *r*^{2}

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:

*x*^{2} + *y*^{2} = 1^{2}

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.

*x*^{2} + *y*^{2} = 1^{2}

(*r* cos ɵ)^{2} + (*r* sin ɵ)^{2} = 1^{2}

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

*r*^{2}(cos ɵ)^{2} + *r*^{2}(sin ɵ)^{2} = 1^{2}

*r*^{2}(cos^{2 }ɵ + sin^{2 }ɵ) = 1^{2}

But hey, we already decided that *r* = 1.

1^{2}(cos^{2 }ɵ + sin^{2 }ɵ) = 1^{2}

cos^{2 }ɵ + sin^{2 }ɵ = 1^{2}

Since 1^{2} equals 1, we've finally made it:

cos^{2 }ɵ + sin^{2 }ɵ = 1

High fives all around.

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:

sin^{2} ɵ + cos^{2 }ɵ = 1

Now just plug in what we know.

(0.65)^{2} + cos^{2 }ɵ = 1

0.4225 + cos^{2 }ɵ = 1

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

cos^{2 }ɵ = 1 – 0.4225

cos^{2 }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

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

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.

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:

Given in Quadrant III, find cos ɵ.

Make a sketch like this:

*x* = ? *y* = -91

r

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:

### Sum and Difference Identities

First, allow us to introduce you to the**sine sum and difference identities**.sin(α + β) = sin(α)cos(β) + cos(α)sin(β)

and

sin(α – β) = sin(α)cos(β) – cos(α)sin(β)

Pretty wild, right? Let's prove that they actually work.

Take a look at the figure below, which shows two angles α and β.

Imagine Line

*OB*is swinging around*O*and sweeps out the angle α and then sweeps out the angle β. (Things are definitely moving here in trig land.)Look carefully, and you'll see that perpendicular lines have been added to form a few right triangles.

First off, notice that ∠

*BOE*= α + β.This means that sin(∠

*BOE*) = sin(α + β).Now, look at Δ

*OAE*and bust out our trig ratio for the sine.Since

*BD*=*AC*, we can rewrite that as .Stay with us now…it's time to do some rearranging.

Let's shuffle the denominators. (Poker, anyone?)

Hold onto that equation for a sec while we take a quick side trip. We need to show that ∠

*CED*is the same as angle α for the rest of the proof to work.Okay, see how the line

*OD*crosses both the parallel lines*CD*and*OB*? That means ∠*ODC*has the same measure as angle α because of the alternate interior angles theorem from back in Geometry. We can also see from the drawing that ∠*ODC*and ∠*CDE*form a right angle, so m∠*CDE*= 90° – m∠*ODC*. And since ∠*ODC*= α, we can rewrite that as m∠*CDE*= 90° – α. Right? Right.Since the three angles in triangle

*CDE*are gonna add up to 180° (like any triangle), that means:m∠

*DCE*+ m∠*CDE*+ m∠*CED*= 180°90° + (90° – α) + m∠

*CED*= 180°180° – α + m∠

*CED*= 180°-α + m∠

*CED*= 0m∠

*CED*= αWhew. Good deal; that angle way up at the top of the figure has the same measure as angle α. Trust us, that's important.

Now we'll look back at the diagram one last time and write down a bunch of trig ratios for the angles α and β.

(This one works because of all that work we just did to show that m∠

*CED*= α.)How do those guys help us? Welp, we can now substitute all those gnarly ratios we came up with earlier:

**sin(α + β) = sin(α)cos(β) + cos(α)sin(β)**Victory! That's our

**sine sum identity**.Now, we can find the difference identity by plugging in -β for β like this:

sin(α – β) = sin[α + (-β)] = sin(α)cos(-β) + cos(α)sin(-β)

From our negative angle identities, we know that:

cos(-β) = cos(β)

and

sin(-β) = -sin(β)

Which means we have our

**sine difference identity**:**sin(α – β) = sin(α)cos(β) – cos(α)sin(β)**### Cosine Gets a Turn

Now onto the

**cosine sum identity**.**cos(α + β) = cos(α)cos(β) – sin(α)sin(β)**We can prove the cosine sum identity by applying our co-function identity:

cos(α + β) = sin[90° – (α + β)]

Remove those pesky parentheses:

cos(α + β) = sin(90° – α – β) = sin[(90° – α) – β]

Next, plug this into our sine difference identity:

cos(α + β) = sin[(90° – α) – β] = sin(90° – α)cos(β) – cos(90° – α)sin(β)

Using co-function identities again, we can swap out sin(90° – α) for cos(α), and swap cos(90° – α) for sin(α) to finish up.

cos(α + β) = cos(α)cos(β) – sin(α)sin(β)

Bam. Nailed it.

To find the cosine difference identity, plug in -β for β, just like you did for the sine difference identity.

cos(α – β) = cos[α + (-β)] = cos(α)cos(-β) – sin(α)sin(-β)

Sub in those negative angle identities to get the

**cosine difference identity**:**cos(α – β) = cos(α)cos(β) + sin(α)sin(β)**Now let's take our hard-earned sum and difference identities, and use them to solve problems.

### Sample Problem

Use a sum or difference identity to find the exact value of cos(75°) without a calculator.

To work this, we look at the 75° to see if it's the sum or difference of any angles from our reference triangles.

We see that 75° = 30° + 45°.

So:

cos(75°) = cos(30° + 45°)

We can use the cosine sum identity.

cos(α + β) = cos(α)cos(β) – sin(α)sin(β)

cos(30° + 45°) = cos(30°)cos(45°) – sin(30°)sin(45°)

Now plug in values from your reference triangles. (Oh, it's them again.)

And the answer is:

### Sample Problem

Use a sum or difference identity to find the exact value of sin(165°).

What quadrant is 165° in?

It's bigger than 90° but smaller than 180°, so it's in Quadrant II.

To work this, we look at the 165° to see if it's the sum or difference of our angles from our reference triangles.

165° = 120° + 45°

So:

### Sample Problem

Use a sum or difference identity to find the exact value of cos(255°).

What quadrant is 255° in?

Yup, it's in Quadrant III. Let's look at the 255° to see if it is the sum or difference of any special angles.

255° = 300° – 45°

So:

### Going Off on a Quick Tangent

The tangent sum and difference identities can be found from the sine and cosine sum and difference identities. Lucky for us, the tangent of an angle is the same thing as sine over cosine.

Plug in the sum identities for both sine and cosine.

Next, a little division gets us on our way (fractions never hurt).

Divide the numerator and the denominator by cos(α)cos(β).

Now, split up the terms.

Replace each term with "1" or "tan," wherever appropriate. Remember, tangent is sine over cosine.

Now simplifying just a little more gives us the

**tangent sum identity**:Next, we can find tan(α – β) the same way.

Once again, with a little help from our friend division, divide the numerator and the denominator by cos(α)cos(β).

Now, split up the terms and replace each term with "1" or "tan," wherever appropriate.

We're almost there—there

*is*a there, we promise. Simplifying just a little more will do the trick. Here's our**tangent difference identity**:### Sample Problem

Use a sum or difference identity to find tan(255°).

First, rewrite 255° as a sum or difference.

255° = 300° – 45°

Which means

We have to clean up our denominator (so far, it's looking

*u-u-u-ugly*).So multiply by the conjugate over the conjugate:

And our answer is:

We've made out it of trig land—alive (or barely alive). See, the world of angles, triangulation and all, isn't so bad.

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