INSTRUCTOR: 5.2 Verifying Trigonometric Identities. Verifying a trigonometric identity means to prove that the two sides of the equation are equal to each other. So here are some helpful strategies to verify trig identities. 

First of all, you need to learn the fundamental identities. And then whenever you see either side of a fundamental identity, the other side should come to mind. Also, be aware of equivalent forms of the fundamental identities. 

And then you want to try to rewrite the more complicated side of the equation so that it's identical to the simpler side. It's sometimes helpful to express all trigonometric functions in the equations in terms of sine and cosine and then simplify the result. 

Usually, any factoring or indicated algebraic operations should be performed. These algebraic identities are often used in verifying trigonometric identities. When selecting substitutions, keep in mind the side that's not changing because it represents the goal. We don't want to change both sides. 

If an expression contains 1 plus sine of x, multiplying both the numerator and the denominator by 1 minus sine of x would give 1 minus sine squared theta or x, which could be replaced with cosine squared theta. Similar procedures apply for 1 minus sine of x, 1 plus cosine of x, and 1 minus cosine of x. 

So let's verify the following equations. If we look at A, we have cotangent of theta divided by cosecant of theta equals cosine of theta. So the more simpler side, of course, would be the right side. 

So let's try to simplify the left side to look like the right side of the equation. And following one of the suggestions or the helpful strategies, let's get the left side in terms of sine and cosine. 

All right, well, cotangent is equal to cosine of theta over sine of theta. And then cosecant of theta is equal to 1 over sine of theta. All right, so remember we're trying to get it to look like the right side. 

Well, if we take the left side, we have a fraction divided by a fraction. So we're going to multiply by the reciprocal, cosine of theta over sine of theta times-- we're going to flip the denominator. So we have sine of theta over 1. OK, so the sine of theta is cancel, which gives us cosine of theta over 1, which is just cosine of theta. 

So once we get both sides to look the same, then we're done. In B, we have 1 minus sine squared theta. So if we think of our Pythagorean identity, if we just subtract sine of theta from both sides, we know that this is equal to cosine squared theta. Excuse me. 

And, of course, this left side was more complicated than the right side. So we're trying to get the left side to look like the right side. So we have two cosine of thetas on top and one on the bottom, so one of those will cancel. And we're just left with cosine of theta. And we're done. 

All right, here in C, we have cosine squared theta times tangent squared theta plus 1 equals 1. Well, of course, the right side is the simpler side. So we want to try to get the left side to look like the right side. 

Well, that tangent squared theta plus 1, if we look at our Pythagorean identities, that is equal to secant squared theta. So I'm going to replace tangent squared theta plus 1 with secant squared theta. 

Now I know that I can change this secant squared theta into terms of cosine by writing it as its reciprocal. So secant squared theta is equal to 1 over cosine squared theta. And these would cancel, and we get 1 equals 1. 

All right, a lot of times, one of the strategies if we have a complex fraction, we can separate it into two separate fractions. Sometimes when it's two separate fractions, we can combine it into one. So this sine of theta minus cosine of theta over sine, we can separate that into two separate fractions. And that does look like the more complicated side. So let's do that. 

So we have sine of theta over sine of theta minus cosine of theta over sine of theta. Well, sine of theta divided by sine of theta is 1. And cosine of theta divided by sine of theta is cotangent of theta. And now the two sides equal. 

Here in E, the more complicated side is going to be the left side. And if I look at the different trig functions or values that I have, we have cosine, and we have cosine. And, of course, we're trying to get it to look like the right side of the equation. So on the right side of the equation, we just have cosine. 

So if I think about sine squared theta in my Pythagorean identity, I can get this into terms of cosine because sine would be equal to 1 minus cosine squared theta based on our Pythagorean identity. And, of course, that's over 1 plus cosine of theta. 

Now we can apply the same rules and methods that we deal with in algebra to deal with trig functions. So you always have to keep in mind what your end goal is. 

And, of course, we wouldn't want to go backwards and change the 1 minus cosine squared theta to sine squared theta. But if we look at this, 1 is a perfect square, cosine squared theta is a perfect square, and it's subtraction. 

So we can use difference of perfect squares to factor this. So this would factor into 1 plus cosine squared theta. Not squared. Sorry, 1 plus cosine of theta, 1 minus cosine of theta. And that's over 1 plus cosine of theta. 

And so if it's a binomial, if the whole binomial is the same, it can cancel out. So those would cancel. And I'm left with 1 minus cosine of theta, which is what the other side is. 

All right, let's look at this last one. We have 1 plus 2 sine of theta cosine of theta over sine of theta plus cosine of theta squared equals 1. So, of course, the left side is more complicated than the right side. 

Well, in E, we factored to get things simpler to reduce. Well, in F, this would be considered to be factored. This is sine of theta plus cosine of theta times sine of theta plus cosine of theta because it's squared. 

So sometimes it helps if we multiply things together. So let's do that. I'm going to keep the numerator the same. And then in the denominator, we're going to multiply this out. 

And hopefully, eventually, we're going to get it all to equal 1. So if I multiply-- let me rewrite the numerator. So if I multiply this out, I'm foiling. 

Sine of theta times sine of theta is sine squared theta. Sine of theta times cosine of theta is just sine theta cosine theta. And then I get the same thing in the middle. 

So let me go ahead and put those together. Sine of theta cosine of theta plus sine of theta cosine of theta would give me 2 sine of thetas cosine of thetas. And then cosine of theta times cosine of theta would be cosine squared theta. 

All right, well, if we think about our Pythagorean identity or identities, sine squared theta plus cosine squared theta equals 1. And since we're adding, we can add in any order. 

So if I add those two together, that would equal 1. So that gives me 1 plus 2 sine theta cosine theta on the bottom. And that's what I have on top also. 

And, of course, if we have something divided by itself, it equals 1. So we have the same thing on the top and on the bottom, so that's just going to be 1. And that's how we simplify trig identities or trig functions.