Hey everyone. As you continue to work through problems dealing with trig expressions, you may come across an expression that looks like this one. The sine squared of 11 pi over 6 plus the cosine squared of 11 pi over 6. Now if this is the first time that you're seeing an expression with multiple squared trig functions, this can look really intimidating. But what if I told you that this was just equal to 1? Well, here I'm going to show you exactly how to get there using what's called the Pythagorean identities, which we are going to use specifically to simplify trig expressions with squared trig functions. Now here I'm going to walk you through exactly where these Pythagorean identities come from and how to use them no matter what form they're in. And soon you'll be able to look at expressions such as this one and know that it's equal to 1 without a second thought. So let's go ahead and get started.

Now, it probably won't come as a surprise that our Pythagorean identities are derived from using the Pythagorean theorem, which you're probably really familiar with. So looking at our triangle here and setting up my Pythagorean theorem a squared plus b squared equals c squared. If I plug in my side lengths of y and x and my hypotenuse of 1 I end up with y squared plus x squared is equal to 1 squared, which I know is just equal to 1. Now with this equation here, I can actually get even more specific with these side lengths because taking a deeper look at my triangle here with this angle of pi over 6 and this hypotenuse of 1, using my knowledge of the unit circle, we already know that this side length of x is actually equal to the cosine of pi over 6 and my side length of y is equal to the sine of pi over 6.

So plugging these values into my equation over here, I end up with sin2π6 plus cos2π6 is simply equal to 1. But you don't have to take my word for it that this is equal to 1 because we know these values. Right? The sine of pi over 6 is equal to 1/2. And the cosine of pi over 6 is equal to the square root of 3 over 2. So taking those values and squaring them, I end up with 1 fourth plus 3 fourths. Now 1 fourth plus 3 fourths is equal to 4 over 4, which we know is just equal to 1. So the sine squared of pi over 6 plus the cosine squared of pi over 6 is indeed equal to 1.

But this doesn't just work for this angle because we can actually generalize this for any angle and this gives us our first Pythagorean identity, that the sine squared of an angle theta plus the cosine squared of an angle theta is simply equal to 1. Now we can also derive our other 2 Pythagorean identities by simply taking this first identity and dividing it by either cosine or sine. So if I were to divide this first identity by cosine, I end up with my second Pythagorean identity that the tangent squared of theta plus 1 is equal to the secant squared of theta. And then if I instead divided that first identity by the sine, I would end up with 1 plus the cotangent squared of theta is equal to the cosecant squared of theta.

And looking at my expression over here, the sine squared of 11 pi over 6 plus the cosine squared of 11 pi over 6, seeing as these angles are the same, just like I see in my identity here, I know that this is simply equal to 1. Now we're going to want to use our Pythagorean identities whenever we see trig functions that are squared, and we're not always just going to be evaluating expressions with angles in them. Sometimes we're going to be asked just to rewrite trig expressions and in order to do that, we're going to have to recognize different forms of these Pythagorean identities. Now this can be tricky the first time that you do it, so let's go ahead and walk through some examples together.

In this first example here, I have the secant squared of theta minus the tangent squared of theta. And here we're asked to use our Pythagorean identities to rewrite the expression as a single term here. So here we don't have any angles, just variables, and we want to rewrite this as a single term. Now looking at this expression here, the secant squared of theta minus the tangent squared of theta, it might not immediately be apparent how we can simplify this using our Pythagorean identities. But looking up at our Pythagorean identities here, I have this second one, the tangent squared of theta plus 1 is equal to the secant squared of theta plus 1. Now these have the same 2 trig functions that I see in my example here. So let's go ahead and see what we can do with this trig identity. The tangent squared of theta plus 1 is equal to the secant squared of theta. Now looking at this knowing what my goal is here if I were to go ahead and subtract the tangent squared of theta from both sides it would end up canceling on that left side just leaving me with 1 and then on that right side I'm left with the secant squared of theta minus the tangent_squared of theta. Now this is the exact expression that we were