Introduction to Trigonometric Identities - Online Tutor, Practice Problems & Exam Prep

Understanding the even and odd identities of trigonometric functions is crucial for simplifying expressions. The cosine function is even, meaning cos(-θ)=cos(θ), while the sine function is odd, so sin(-θ) = -sin(θ). Pythagorean identities, such as sin²(θ) + cos²(θ) = 1, help in simplifying complex trigonometric expressions and verifying identities effectively.

Even and Odd Identities

Hey, everyone. You may remember learning that some functions can be classified as being either even or odd based on the symmetry of their graph. Well, here we're going to specifically take a look at our trig functions and whether they're even or odd. Now, why that's useful might not be immediately apparent to you. But if we know that some function f(x) is even or odd, then we can easily find and simplify f(-x), which is something that we'll have to do more and more as you continue to work your trig problems. So here, I'm going to break down for you exactly which trig functions are even and which are odd and then use that in order to simplify some expressions. So let's go ahead and get started.

Now, let's start by taking a look at our cosine function and specifically the value of our cosine function at π2. Now, here on my graph, I see that f(π2) is equal to 0 and if I go to the other side of my graph and check out the value of my function for -π2, I see that f(-π2) is also equal to 0. So relating these two points to each other, I could say that f(-π2) is equal to f(π2) because they are the exact same value. But I can actually generalize this for the entire function. See f(-x) for this function is always going to be equal to f(x). So if I flip the sign of my input, like I saw here for -π2, the sign of my output remains the same. And that's because my cosine function is an even function.

Now, you might also hear even functions talked about in terms of their symmetry. And even functions are specifically symmetric on the y-axis, which just means that if I took my graph and folded it along that y-axis, all of my points would match up with each other on either side because it's symmetric about that line. Now we've seen that our cosine function is even. But let's take a look at our sine function over here.

Now, looking at our sine function again at a value of π2, I see that f(π2) is equal to positive 1. Then going to the other side of my graph, looking at f(-π2), I see that f(-π2) is actually equal to negative one. So relating these two points to each other, I could say that f(-π2) is equal to -f(π2) because my value for f(-π2) was negative 1, and for positive π2 was positive 1. Now again, we can generalize this for our entire function. So f(-x) here, we see, is always going to be equal to -f(x). So, whenever we flip the sign of our input value to -

Example 1

We just saw how to use our even-odd identities in order to rewrite expressions with negative arguments, and we specifically saw our arguments with negative angle measures in them. But you're actually more often going to have to simplify expressions that just have variables rather than angle measures. So let's take a look at that here. Here we're asked to use our even-odd identities in order to rewrite the expression with no negative arguments in terms of just one trigonometric function. So here, in our first example, we have the negative tangent of negative theta. Now, this is already in terms of just one trigonometric function, but we need to get rid of that negative argument there. I know that the tangent of negative theta is equal to negative tangent of theta. So, I can simply replace the tangent of negative theta using that identity. I want to make sure that I'm paying attention to everything that I need to keep here, so make sure that you're still keeping that negative on the outside. Then the tangent of negative theta, I can replace that with negative tangent of theta. Now I see here that I have two negative signs and I know that two negatives make a positive, so this ends up just being a positive tangent of theta. And now, we have successfully rewritten this with no negative arguments in terms of just one trigonometric function. So we're done with that example there. Let's move on to our second example. Here we're asked to rewrite this expression, the sine of negative theta over the cosine of negative theta. Now here I can go ahead and use my even-odd identities for sine and cosine. And I know that the sine of negative theta is equal to negative sine of theta. Then the cosine of negative theta, because it's an even function, is simply equal to the cosine of theta. Now from here, we have gotten rid of those negative arguments, but it's still in terms of two trig functions. So how can we simplify this further? Well, the sine over the cosine is simply the tangent. So keeping that negative sign, this ends up just being negative tangent of theta - sin ( θ ) cos ( θ ) and that's my final answer here. Now you might have seen a different way to do this and that's totally fine. There are always multiple ways to simplify. Then using our even-odd identity for tangent, we know that the tangent of negative theta is simply equal to negative tangent of theta, so we would end up getting that exact same answer, negative tangent of theta. Now it doesn't matter which way you choose to simplify this because you'll get the right answer regardless. Now that we've seen how to do this using variables, let's continue practicing. Thanks for watching, and I'll see you in the next one.

Use the even-odd identities to evaluate the expression.

$\cos\left(-\theta\right)-\cos\theta$

• A. 0
• B. $-\cos\theta$
• C. $2\cos\theta$
• D. $-2\cos\theta$

Use the even-odd identities to evaluate the expression.

$-\cot\left(\theta\right)\cdot\sin\left(-\theta\right)$

• A. $\tan\theta$
• B. $-\cos\theta$
• C. $\cos\theta$
• D. $\frac{\cos\theta}{\sin^2\theta}$

Select the expression with the same value as the given expression.

$\sec\left(-\frac{4\pi}{5}\right)$

• A. $\cos\left(\frac{4\pi}{5}\right)$
• B. $-\cos\left(\frac{4\pi}{5}\right)$
• C. $\sec\left(\frac{4\pi}{5}\right)$
• D. $-\sec\left(\frac{4\pi}{5}\right)$

Select the expression with the same value as the given expression.

$\sin\left(-38\degree\right)$

• A. $\sin38\degree$
• B. $-\sin38\degree$
• C. $-\sin\left(-38\degree\right)$
• D. $\frac{1}{-\sin38\degree}$

Pythagorean Identities

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

Example 2

Hey, everyone. In this example problem, we're asked to use our Pythagorean identities in order to rewrite the expression with no fraction. And the expression that we're given here is \( \frac{1}{1 + \cos(\theta)} \). Now it might not be immediately apparent how we can use our Pythagorean identities here because we don't even have a squared trig function, but let's play around with this a little bit. In my denominator, I have \( 1 + \cos(\theta) \). Now let's see what happens if I multiply this by \( 1 - \cos(\theta) \). Now remember we're working with a fraction, so if I'm multiplying the denominator let's see what happens when I multiply this out here. In my numerator, I of course just have \( 1 - \cos(\theta) \), having multiplied that by 1. But in my denominator, \( 1 + \cos(\theta) \times 1 - \cos(\theta) \), using my difference of squares, I get that this is \( 1 - \cos^2(\theta) \). Now I have a squared trig function and I can recognize this as being just a different form of that first Pythagorean identity. So if I subtract \( \cos^2(\theta) \) from both sides of this Pythagorean identity it leaves me with \( \sin^2(\theta) = 1 - \cos^2(\theta) \), which is the exact expression that I have here. So this is a sort of trick if you want to get a Pythagorean identity when you don't have one. If you have \( 1 + \) some trig function, if you multiply it by \( 1 - \) that same trig function, you'll end up with a Pythagorean identity. Now using that here, I can replace that denominator with \( \sin^2(\theta) \). And in my numerator, I still just have that \( 1 - \cos(\theta) \). But remember what our goal was here, we don't want a fraction. So how can we get rid of this fraction? Well let's think about this. In this expression, I could rewrite this as being \( \frac{1}{\sin^2(\theta)} \times (1 - \cos(\theta)) \). And now the only reason that we have a fraction here is this \( \frac{1}{\sin^2(\theta)} \). But remember, \( \frac{1}{\sin(\theta)} \) is just the cosecant. So \( \frac{1}{\sin^2(\theta)} \) is simply the cosecant squared of theta. So now I'm just left with \( \csc^2(\theta) \times (1 - \cos(\theta)) \). No more fraction. So this is my final answer here that \( \frac{1}{1 + \cos(\theta)} = \csc^2(\theta) \times (1 - \cos(\theta)) \). Let me know if you have any questions. Thanks for watching, and I'll see you in the next one.

Use the Pythagorean identities to rewrite the expression as a single term.

$\left(1+\csc\theta\right)\left(1-\csc\theta\right)$

• A. 1
• B. $-\csc^2\theta$
• C. $\cot^2\theta$
• D. $-\cot^2\theta$

Use the Pythagorean identities to rewrite the expression with no fraction.

$\frac{1}{1-\sec\theta}$

• A. $1+\sec\theta$
• B. $\frac{1}{\tan^2\theta}$
• C. $-\cot^2\theta\left(1+\sec\theta\right)$
• D. $-\tan^2\theta\left(1+\sec\theta\right)$

Simplifying Trig Expressions

Hey, everyone. We just became more familiar with our even, odd, and Pythagorean trig identities, and we'll now be asked to simplify some more complicated trig expressions. So, we'll have to use these new identities along with all of those identities that we already know really well in order to fully simplify these expressions. Now, simplifying trig expressions can be tricky because there's not just one way to get to the correct answer, and there are also not specific steps that you can follow in order to get it right every single time. But here, I'm going to break down for you exactly what it means for a trig expression to be fully simplified and then walk you through some strategies to help you get there. So let's go ahead and get started by jumping right into our first example where we're asked to, of course, simplify the expression.

Now, the expression that we're given here is the tangent of negative theta times the cosecant of theta. As we're simplifying these trig expressions, it's important that we know what it means for a trig expression to be fully simplified. There are three things that tell us a trig expression is fully simplified. The first of these is that all of our arguments must be positive. Looking at my expression, I see that I'm taking the tangent of negative theta. So how can we fix that? Well, looking at our strategies, our very first strategy, and probably the most important one, is that we want to be constantly scanning for identities. Since I have the tangent of negative theta, that should tell me that I need to take a look at my even-odd identities, which tell me that the tangent of negative theta is equal to the negative tangent of theta. So using that identity, I can rewrite this to have a positive argument. The tangent of negative theta using that identity becomes the negative tangent of theta. Now my expression is the negative tangent of theta times the cosecant of theta, and now all of my arguments are positive.

Now let's take a look at the two other things that tell us a trig expression is fully simplified. The first is that our expression should contain no fractions, which it currently doesn't. And the next one is that our expression should contain as few trig functions as possible. Looking at my expression, I have two trig functions. I have the tangent and I have the cosecant. So how can we get that to be fewer trig functions? Well, taking a look at our strategies, we are still constantly scanning for identities. And our next strategy tells us that we should break down everything in terms of sine and cosine. Remember, I have two trig functions: the tangent and the cosecant. Taking a look at my identities, I know that the cosecant is equal to 1 over sine and the tangent is equal to sine over cosine. So, I can use those two identities in order to break this down into just terms of sine and cosine. The negative tangent of theta becomes negative sine of theta over the cosine of theta. Then the cosecant of theta is just 1 over the sine of theta. Having broken this down in terms of sine and cosine allows me to cancel some stuff out. Because I have sine on the top and sine on the bottom, that goes away, and I am just left with negative one over the cosine of theta.

So now I have just one trig function. But I do have a fraction now. And remember, to be fully simplified, our expression should contain no fractions. But remember, we are constantly scanning for identities. And looking at my identities, I know that one over the cosine is simply the secant. So this negative one over the cosine that I have in my expression can be simplified down to the negative secant of theta. Now in this expression, all of my arguments are positive, my expression contains no fractions, and it has as few trig functions as possible. So, the tangent of negative theta times the cosecant of theta fully simplifies down to the negative secant of theta. Now, you may have noticed here that we didn't use all of our strategies, and that's totally okay. Sometimes you're not going to need all of them, but we're going to continue using more in order to get where we want to go.

So let's go ahead and take a look at another example and see if we can use some more strategies. In this next example, we are still simplifying the expression, but the expression that we're given here is the sine squared of theta over 1 plus the cosine of theta. Now remember, for this trig expression to be fully simplified, all of our arguments need to be positive, which they already are here. Our expression should contain no fractions, and we want as few trig functions as possible. Now this expression is just one big fraction. So let's figure out how to get rid of that. Now looking at our strategies, remember we are constantly scanning for identities. We want to break down everything in terms of cosine and sine. But this entire expression is already in terms of cosine and sine, so that's not really going to help me. Now taking a look at our fourth strategy here, if we have 1 plus or minus some trig function, we want to multiply both the top and the bottom of our fraction by 1 minus or plus that same trig function. Now looking at my expression, I have 1 plus the cosine of theta. So using that strategy, I want to multiply this by 1 minus the cosine of theta, having the opposite sign of what my original expression had. But remember, if I'm multiplying the bottom by that, I also need to multiply the top as not to change the value of the expression.

Now multiplying this in my numerator, I'm going to leave that factored as the sine squared of theta times 1 minus the cosine of theta. Now in my denominator, I recognize this as being a difference of squares because I have the same two terms, 1 and cosine, just one with a plus and one with a minus. So multiplying this out gives me 1 minus the cosine squared of theta. Now from here, remember we're constantly scanning for identities. And taking a look at my identities here, I know that the sine squared of theta plus the cosine squared of theta is equal to 1. And if I subtract the cosine squared of theta from both sides here, it cancels on that left side, leaving me with exactly what's in the denominator of my expression here, 1 minus the cosine squared of theta. Now I know that 1 minus the cosine squared of theta is equal to the sine squared of theta. So I can replace my denominator. Now in my numerator, I'm keeping that the same for now, the sine_squared of theta times 1 minus the cosine of theta. But in my denominator, I am using that Pythagorean identity in order to replace that with the sine squared of theta. Now I have the sine squared in both the numerator and the denominator, and that cancels out. Now all I'm left with is 1 minus the cosine of theta.

So let's figure out if we need to do anything else here. Here we see that all of our arguments are positive, our expression now contains no fractions, and we have just one trig function, so as few trig functions as possible. We have now successfully fully simplified this down from the sine squared of theta over 1 plus the cosine of theta down to just 1 minus the cosine of theta. Now remember, there are multiple ways to successfully simplify an expression. So, if you want to do this differently, feel free to do whatever works for you. Now that we've seen how to simplify some trig expressions, let's continue practicing together. Thanks for watching, and I'll see you in the next one.

Example 3

Hey, everyone. In this problem, we're asked to simplify the expression. And the expression we're given is the sine squared of theta minus the tangent squared of theta over the sine of theta plus the tangent of theta. Now remember that in order to fully simplify your equation, we want all of our arguments to be positive, which here, they already are. And we also don't want any fractions, which here, we do have one big fraction, so we want to get rid of that first. Now looking at my strategies here, we are constantly scanning for identities, but I don't really see any that will help me yet. But one of my other strategies is going to be to factor. And looking at the numerator that I have here, sine squared of theta minus the tangent squared of theta, you may recognize this as being a difference of squares. Because this is one term squared minus another term squared, so this will factor to that first term plus that second term times that first term minus that second term. So factoring that out, I can get the sine of theta, that first term, plus the tangent of theta, the second term, and then multiplying that by the sine of theta minus the tangent of theta. Now my denominator stays the same here, the sine of theta plus the tangent of theta. But here I notice that I can cancel something because this entire denominator is exactly what this first term now is in my numerator. So this fully cancels out and I've gotten rid of my fraction because now all I'm left with is the sine of theta minus the tangent of theta. So I have no fractions here. Now the other thing that tells us our expression is fully simplified is that we have as few trig functions as possible. Now here I have 2 trig functions but because they're being subtracted, it's going to be kind of hard to make it so that there are any fewer trig functions than are already there. So having these 2 trig functions is as few trig functions that I can have as possible. So we have also satisfied that last piece of criteria there, and we have fully simplified our trig expression down to the sine of theta minus the tangent of theta. Thanks for watching and let me know if you have any questions.

Example 4

Hey, everyone. In this problem, we're asked to simplify the expression. And the expression that we're given is the cosine of theta plus the cosecant of theta divided by the cosine of theta plus the sine of theta minus the secant of theta over the sine of theta. Now we want to fully simplify this expression. And remember that for an expression to be fully simplified, we want all of our arguments to be positive, which here they already are. We want no fractions and we want as few trig functions as possible. So let's go ahead and take a look at our strategies here.

Now here I have two fractions that are being added together, and one of my strategies tells me to go ahead and add any fractions using a common denominator here. So looking at these two fractions, what is our common denominator going to be? Well, in this first fraction, I have a denominator of cosine, and in my second fraction, I have a denominator of sine. So what is our common denominator going to be here? Well, I can multiply these two things together in order to give me a common denominator of the cosine times the sine. Now, in order to do that, to get that common denominator and add these two fractions together, remember that we need to multiply the top and the bottom of each fraction in order to get that common denominator. So here I'm going to multiply this first fraction by the sine of theta over the sine of theta. And then I'm going to multiply my second fraction by the cosine of theta over the cosine of theta.

Now looking at this, multiplying that bottom is going to give me that common denominator there. But we need to multiply at the top as well. So we need to go ahead and distribute that sine into these two terms here, starting off my denominator with sine of theta times cosine of theta plus the sine of theta times the cosecant of theta. So that is my first fraction fully multiplied, giving me that common denominator. Then for my second fraction, I need to do the same thing. I need to distribute that cosine into both of my terms. Now that first term will give me plus the cosine of theta times the sine of theta and then minus, because that's a subtraction right there, minus the cosine of theta times the secant of theta.

Okay. This fraction looks a little bit crazy right now, but let's look back to our strategies. Now remember, we want to be constantly scanning for identities, and something that can be helpful is to break down in terms of sine and cosine. Now looking down to my identities here I know that the cosecant is equal to 1 over sine, and the secant is equal to 1 over cosine. So I can break this entire thing down in terms of sine and cosine by using those identities to replace the cosecant and the secant in this expression. So let's go ahead and do that here. Now everything is staying the same besides the cosecant and the secant. So I still have that first term, sine of theta times cosine of theta, but now my second term is sine of theta times one over the sine of theta using that cosecant reciprocal identity. Then that other term will stay the same, cosine of theta times the sine of theta. And then this term becomes cosine of theta times 1 over the cosine of theta.

Now remember we still have a fraction, denominator here, and it is still the cosine of theta times the sine of theta. Now the only thing that I changed was this to 1 over sine and this to 1 over cosine. But let's look at what happens here. Now here I have a sine on the top and I have it being multiplied by 1 over the sine. So those are going to cancel out. Now when those cancel out that's just going to leave me with a value of positive 1. Then over here this cosine cancels with this cosine and I'm left with a negative one. So here I have a positive one and I have a negative one. What happens when I take 1 and subtract 1? I get 0. So those fully go away, canceling each other out. Now all that I'm left with in that numerator is sine theta times cosine theta plus cosine theta times sine theta. Those are the only two terms that I have left in that numerator with everything else having canceled out.

Now we do still have a fraction, so my denominator is still the cosine of theta times the sine of theta. But let's make sure that we take a break and we pause and we look at what's actually happening here. I have the sine of theta times the cosine of theta plus the cosine of theta times the sine of theta. Now those are the same two trig functions being multiplied together so these are actually the same exact term. So if I take one of those terms and I add it to another one of those terms, how many of those terms do I have? I have 2. So this is really just 2 times the sine of theta times the cosine of theta in that numerator. Now this is still being, of course, divided by the cosine of theta times the sine of theta. But, again, let's look at what's really happening here. I have a sine on the top. I have a sine on the bottom. I have a cosine on the top and I have a cosine on the bottom. So what's the only thing that I'm left with? 2. And that's my final answer here because I don't have any fractions and I have as few trig functions as possible because I have literally no trig functions. So our original expression here, this long mess of fractions, is really just equal to 2. So I know that that was a lot. So let me know if you have any questions. Thanks for watching, and I'll see you in the next one.

Simplify the expression.

$\tan^2\theta-\sec^2\theta+1$

• A. 0
• B. 1
• C. $\csc^2\theta+1$
• D. 2

Simplify the expression.

$\frac{\tan\left(-\theta\right)}{\sec\left(-\theta\right)}$

• A. $\sin\theta$
• B. $-\sin\theta$
• C. $-\cot\theta$
• D. 1

Simplify the expression.

$\left(\frac{\tan^2\theta}{\sin^2\theta}-1\right)\csc^2\left(\theta\right)\cos^2\left(-\theta\right)$

• A. $\cot^2\theta$
• B. $\tan\theta$
• C. 1
• D. – 1

Verifying Trig Equations as Identities

Hey, everyone. Now that we know how to simplify trig expressions, you're going to come across a new type of problem in which you're given a trig equation and asked to verify the identity. This sounds like it could be complicated, but here I'm going to show you that verifying an identity just comes right back down to simplifying, just with the specific goal of both sides of our equation being equal to each other. So here, we're going to keep using our simplifying strategies, just in a slightly different context. And I'm going to walk you through exactly how to do that here. So let's go ahead and get started. In working with these problems, you may only have to simplify one side of your equation or you may have to simplify both sides. This will become more apparent as you begin to work through a problem. Let's go ahead and jump right into our first example here where we're asked to verify the identity. Something that I do want to mention is that sometimes these problems may ask you to prove the identity or establish the identity, but these all mean the same thing.

The identity that we're asked to verify here is sinθcosθ 1−cos2θ is equal to 1tanθ. So how do we start here? Well, in working with these problems, we always want to start by simplifying our more complicated side first. So in looking at this equation, it's clear that this left side is more complicated than that right side. So we're going to start by applying our simplifying strategies to that left side. Looking at my left side, sinθcosθ 1−cos2θ , remember, one of our most important strategies for simplifying is to constantly be scanning for identities. Looking at this left side of my equation this one minus cosine squared of theta looks familiar. And looking at my identities here I know that one of my Pythagorean identities tells me that sin2θ+cos2θ=1. So if I subtract the cosine squared of theta from both sides of this identity, I end up with exactly what's in that denominator on that right side, 1−cos2θ. So using this identity, I can replace that denominator with sin2θ using that Pythagorean identity. Now I'm going to keep my numerator the same as sinθcosθ. And I can do some canceling here. Because in my denominator I have sin2θ and in my numerator I have sinθ. So that sinθ in my numerator goes away. And in my denominator my sin is no longer squared. And I'm simply left with cosθsinθ. Now remember we are still constantly scanning for identities cotθ. Now I also know that the cotangent is equal to 1tanθ, which is exactly what the right side of my equation is. And remember that our ultimate goal is to show that these two sides of our equation are equal to each other. So knowing that this cosine over sine is 1tanθ, I can go ahead and rewrite that side as 1tanθ. Now I don't have to do anything to that right side of my equation. It's just 1tanθ. And now we have that 1tanθ is equal to 1tanθ. So we have successfully verified this identity by showing that the left side of our equation is equal to the right side. Now, even though this is called verifying the identity, this is not a new identity you have to learn. All that we're doing here is showing that two sides of an equation are equal.

Let's go ahead and take a look at our second example here. We're still verifying the identity, but the trig equation that we're given here is sec2θ−tan2θ cos(−θ)+1 is equal to 1−cosθ sin2θ . Now remember that we want to start with our more complicated side first, and sometimes it won't be immediately apparent to you which side is more complicated. That's totally okay. Here, I'm going to start with the left side because this looks like there's a little bit more going on here. But if you were to start with the right side, you could still end up being able to verify this identity. But let's go ahead and get started in simplifying that left side, sec2θ−tan2θ cos(−θ)+1 . Remember here we want to be constantly scanning for identities. And looking at this particular side of my equation, this sec2θ minus tan2θ reminds me of a Pythagorean identity, which tells us that tan2θ+1=sec2θ. So if I rearrange this identity and subtract the tangent squared of theta from both sides, I will end up with exactly what's in that numerator there. And it's simply equal to 1. So I can replace that entire entire numerator with just 1. Then in my denominator, I still have that cos(−θ)+1. Now remember, we're still constantly scanning for identities here. And looking in that denominator, since I have the cosine of negative theta, I know that I can use my even-odd identity in order to get rid of that negative argument here because the cosine of negative theta is simply the cosine of theta. So simplifying that denominator further, leaving my numerator as 1 because that cannot be further simplified, in my denominator, I now just have the cosθ+1. Now this looks rather simplified. So let's go ahead and leave this as is and see if we can get this right side of our equation to be equal to that left side. So let's go ahead and start simplifying here. Here I have 1−cosθ sin2θ . Now, one of my simplifying strategies tells me that if I have one plus or minus a trig function, I want to go ahead and multiply the top and the bottom by 1−+cosθ or plus that same trig function. Here in my numerator, since I have 1−cosθ, I'm going to go ahead and multiply this by 1+cosθ. And remember, if I'm multiplying the top by that, I need to multiply the bottom by that as well to not change the value of our expression. So let's go ahead and multiply this out. Now in my numerator, that is a difference of squares. So this ends up giving me 1−cos2θ. Then in my denominator, I'm going to keep that as is, sin2θ times 1+cosθ. Here in my numerator scanning for identities, this one minus cosine squared of theta is an identity that we've used before, our Pythagorean identity. Now I know that one minus the cosine squared of theta is simply equal to sin2θ, so I can go ahead and rewrite that numerator as the sin2θ. And then in my denominator, I also have a sin2θ, and that is still multiplied by 1+cosθ. Now here I can cancel some stuff out because I have sin2θ on the top and on the bottom. Now I am simply left with a 1 in that numerator and then a 1+cosθ in that denominator, which is exactly what I have on that left side there. Now I can go ahead and rewrite that left side, just switching the 1 and the cosine, to get them exactly the same, 1+cosθ. And now my two sides of my equation are equal to each other, and we have successfully verified this identity.

Now I know that working through these sorts of problems can be really tricky, so if you're working through one and you just get stuck and you're not quite sure where to go, it can sometimes be a good idea to just completely start over from scratch because you may discover something that you didn't discover before. Just keep getting repetition with these problems and you'll get better and better with every try. Thanks for watching, and I'll see you in the next one.

Example 5

Hey, everyone. In this problem, we're asked to verify the identity by working with one side of the equation. Now looking at this equation that we have, we have 1−sinθ over cosθ − cosθ1+sinθ = 0. So it's very clear what side we should be working with, this left side, because we can't do anything to that 0 on the right. So, let's go ahead and get started and take a look at our strategies here. Now looking at these two fractions that are being subtracted, remember that we want to go ahead and add any fractions using a common denominator. Now, I know these fractions are being subtracted, but subtraction is basically just like adding a negative, so we still want to go ahead and combine these fractions with that common denominator. Now looking at the denominators that I have here, I have the cosθ and 1+sinθ. So my common denominator will take both of these and multiply them together to give me a new denominator of cosθ×(1+sinθ). Now, in order to get that common denominator, I need to go ahead and multiply each of these fractions in order to get that common denominator. Now for that common denominator, I need to multiply this first fraction by 1+sinθ. And if I'm multiplying the bottom by that, I also need to multiply the top by that, 1+sinθ. Now multiplying that out will end up giving me a numerator of 1−sin2θ because this is a difference of squares. Now, let's look at that second fraction. We want to multiply both of these by the cosθ in order to get that common denominator. Now on the top, I am left with that minus cos2θ. Remember that we are subtracting here, so that's why I have that minus. Now where do we go from here? Well, remember, we want to be constantly scanning for identities, and I see a couple of different things happening here. Now in my numerator, I have this 1−sin2θ. And coming over here to my identities, I know that my first Pythagorean identity tells me that sin2θ+cos2θ=1. So if I subtract the sin2θ from both sides, it will cancel on that left side and leave me with that 1−sin2θ, which is exactly what I have highlighted here in my numerator. So that one minus sin squared is equal to the cosine squared of theta. So now in my numerator using that Pythagorean identity, I am left with the cos2θ−cos2θ. And then all of that is over that same denominator, the cosθ×(1+sinθ). Now what's happening here? Well, in my numerator, I have the cos2θ−cos2θ. So what happens when I take something and I subtract that same something? Well, I'm just left with 0. Now that right side of my equation is also 0, so I have successfully verified this identity because 0 is equal to 0 and we're done here. Let me know if you have any questions and thanks for watching.

Example 6

In this problem, we're asked to verify the identity by working with both sides of the equation. The identity that we're given here is the secant of theta times one minus the sine squared of theta is equal to 1 plus the tangent squared of theta times the cotangent squared of theta divided by the cosecant squared of theta times the secant of theta. Now remember that whenever we're working on verifying an identity, we want to start with our more complicated side first, which is very clearly this right side because there is a lot going on there. So let's go ahead and get started and figure out how to simplify this.

Now looking at this right side, 1 plus the tangent squared times the cotangent squared over the cosecant squared times the secant squared. Where can we go from here? Well, remember we want to be constantly scanning for identities and the first thing that I notice here in my numerator is 1 plus the tangent squared. Now 1 plus the tangent squared, taking a look at my Pythagorean identities, is equal to the secant squared. So I can go ahead and replace that in my numerator there. So I have the secant squared of theta times the cotangent squared of theta with the same denominator, the cosecant squared of theta times the secant of theta.

Here I can go ahead and just cancel this secant on the bottom with one of the secants on the top. So now I am left with the secant of theta times the cotangent squared of theta over the cosecant squared of theta. Now, let's take a look at our strategies here. We can go ahead and break down everything in terms of sine and cosine because we can't add any fractions and we don't have 1 plus or minus a trig function and we can't factor here either. So let's go ahead and break this down.

Now I know that the secant of theta is equal to 1 over the cosine of theta and the cotangent squared of theta is equal to the cosine squared of theta over the sine squared of theta. Then in my denominator, I can also rewrite that in terms of sine. It is one over the sine squared of theta. So let's go ahead and work with this and see what we can cancel. Now in that numerator, one of those cosines is going to cancel, and I'm just left with the cosine over the sine squared.

And then looking at this, there's a lot of fractions going on. So let's go ahead and rewrite this so that you can clearly see what I'm gonna do here. So here I have the cosine of theta over the sine squared of theta. Now whenever we divide by a fraction, that's the same thing as multiplying by the reciprocal. So really, I'm multiplying this by the sine squared of theta over 1. Now here I have sine squared on the top, sine squared on the bottom, that goes away. And all I'm left with is the cosine of theta.

Now this looks really simplified. So let's go ahead and work with that left side now to see if we can get that to equal the cosine of theta. Now looking at my left side, the secant of theta times 1 minus the sine squared of theta, let's see what we can do here. We are constantly scanning for identities and looking here I have 1 minus the sine squared of theta, which I know is one of my Pythagorean identities just in a slightly different form because the sine squared of theta plus the cosine squared of theta is equal to 1. So if I go ahead and subtract the sine squared of theta from both sides, I am left with exactly what's in those parentheses there, and that will be equal to the cosine squared of theta.

So I can rewrite this as the secant of theta times the cosine squared of theta. Now where can we go from here? Well, the secant of theta I know is 1 over the cosine of theta. And if this is multiplying the cosine squared of theta, that's going to be really helpful. So what can we do here? Well, I have a cosine on the bottom and a cosine on the top. And since that cosine was squared, I am still left with one of them, and all I'm left with here is the cosine of theta. Now these sides had dramatically different amounts of work that we had to do, but they're still equal to each other. So I can bring this all the way down here because we can clearly see that now this identity is verified. The cosine of theta is equal to the cosine of theta. Thanks for watching, and I'll see you in the next one.

Identify the most helpful first step in verifying the identity.

$\left(\frac{\tan^2\theta}{\sin^2\theta}-1\right)=\sec^2\theta\sin^2\left(-\theta\right)$

• A. Add the terms on the left side using a common denominator.
• B. Rewrite left side of equation in terms of sine and cosine.
• C. Use even-odd identity to eliminate negative argument on right side of equation.
• D. Rewrite right side of equation in terms of sine and cosine.

Identify the most helpful first step in verifying the identity.

$\sec^3\theta=\sec\theta+\frac{\tan^2\theta}{\cos\theta}$

• A. Rewrite left side of equation in terms of sine and cosine.
• B. Subtract $\sec\theta$ from both sides.
• C. Use reciprocal identity to rewrite $\sec\theta$ on right side of equation.
• D. Rewrite $\tan^2\theta$ in terms of sine and cosine.

Here’s what students ask on this topic:

What are the even and odd trigonometric identities?

Even and odd trigonometric identities help simplify expressions involving trigonometric functions. The cosine function is even, meaning $\cos (-\theta )=\cos \left(\theta \right)$. The sine function is odd, so $\sin (-\theta )=-\sin \left(\theta \right)$. Similarly, secant is even, and cosecant, tangent, and cotangent are odd. These identities are useful when dealing with negative arguments in trigonometric functions.

How do you use Pythagorean identities to simplify trigonometric expressions?

Pythagorean identities are essential for simplifying trigonometric expressions. The primary identity is ${\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)=1$. By dividing this identity by ${\cos }^2\left(\theta \right)$ or ${\sin }^2\left(\theta \right)$, you get the other two identities: ${\tan }^2\left(\theta \right)+1={\sec }^2\left(\theta \right)$ and $1+{\cot }^2\left(\theta \right)={\csc }^2\left(\theta \right)$. These identities help in rewriting and simplifying expressions involving squared trigonometric functions.

How do you verify a trigonometric identity?

Verifying a trigonometric identity involves showing that both sides of an equation are equal. Start by simplifying the more complicated side using known identities like Pythagorean, even-odd, and reciprocal identities. For example, to verify $\sin \left(\theta \right)\cos \left(\theta \right)/(1-{\cos }^2(\theta \left)\right)=1/\tan \left(\theta \right)$, simplify the left side using the identity $1-{\cos }^2\left(\theta \right)={\sin }^2\left(\theta \right)$. This reduces to $\cos \left(\theta \right)/\sin \left(\theta \right)=1/\tan \left(\theta \right)$, proving the identity.

What are some strategies for simplifying trigonometric expressions?

Simplifying trigonometric expressions involves several strategies: 1) Constantly scan for identities like Pythagorean, even-odd, and reciprocal identities. 2) Break down complex expressions into sine and cosine. 3) Use algebraic techniques like factoring and combining fractions. 4) If you encounter $1+\cos \left(\theta \right)$ or $1-\cos \left(\theta \right)$, multiply by the conjugate. For example, to simplify ${\sin }^2(\theta )/(1+\cos \left(\theta \right))$, multiply by $1-\cos \left(\theta \right)$ to get ${\sin }^2(\theta )/{\sin }^2(\theta )=1$.

How do you determine if a trigonometric function is even or odd?

To determine if a trigonometric function is even or odd, analyze its symmetry. A function is even if $f(-x)=f\left(x\right)$ and odd if $f(-x)=-f\left(x\right)$. For trigonometric functions, cosine and secant are even because $\cos (-\theta )=\cos \left(\theta \right)$ and $\sec (-\theta )=\sec \left(\theta \right)$. Sine, tangent, cosecant, and cotangent are odd because $\sin (-\theta )=-\sin \left(\theta \right)$ and similarly for the others. This property helps in simplifying expressions with negative arguments.