Introduction to Trigonometric Identities - Video Tutorials & Practice Problems

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 of X is even or odd, then we can easily find and simplify F of negative X, which is something that we'll have to do more and more as you continue to work through 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 pi over two. Now, here on my graph, I see that F of pi over two is equal to zero. And if I go to the other side of my graph and check out the value of my function for negative pi over two, I see that F of negative pi over two is also equal to zero. So relating these two points to each other, I could say that F of negative pi over two is equal to F of pi over two because they are the exact same value. But I can actually generalize this for the entire function cf of negative X for this function is always going to be equal to F of X. So if I flip the sign of my input, like I saw here for negative pi over two, 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 I 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, 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 pi over two, I see that F of pi over two is equal to positive one. Then going to the other side of my graph, looking at F of negative pi over two, I see that F of negative pi over two is actually equal to negative one. So relating these two points to each other. I could say that F of negative pi over two is equal to negative F of pi over two because my value for F of negative pi over two was negative one. And for positive pi over two was positive one. Now again, we can generalize this for our entire function. So F of negative X here we see is always go, going to be equal to negative F of X. So whenever we flipped the sign of our input value to negative pi over two, we also had to flip the sign of our output value. And that's because the sine function is an odd function. Now again, you might hear functions being odd talked about in terms of their symmetry. Now, whereas our even functions were symmetric about the Y axis, our odd functions are symmetric about the origin. Now we already saw that our cosine function is an even function and our sine function is an odd function. But what about all of our other trick functions? Well, we already know that C is a one over cosine. So with that in mind, it makes sense that C is also an even function. And we can follow the same logic to find that cosecant is also an odd function because it's just one over sign. But what about tangent? Well, we know the tangent is the sign over the cosine. So here we would be taking an odd function and dividing it by an even function, which actually means that our tangent function will also end up being odd. And by extension cotangent is also odd now that we know whether all of our trig functions are either even or odd. What do we do with that information? Well, all of this information together is collectively referred to as the even odd identities. And as we continue throughout this chapter, it's important for you to know that an identity is just an equation which is true for all possible values in working with different identities depending on your professor or your textbook, you might see identities express with different variables. Now we're going to continue to use the variable theta here when working with our identities. But you may see these written in terms of xyab alpha beta, any number of different variables and just know that no matter what variable that is in your identity, it still means the same exact thing. So here with the cosine of negative, the knowing that our cosine is an even function, we know that the cosine of negative data is simply equal to positive cosine of the. So here with our expression, the cosine of negative pi over four, I know that this is really just equal to the cosine of pi over four. And from here, I can simply use my left hand rule in order to determine what this value is, which is simply equal to the square root of 2/2. So we didn't have to figure out where negative pi over four is on our unit circle because we have our even odd identities. Now taking a look at s of negative data here we know that sine is an odd function. So the sign of negative data is simply equal to negative sine of data. Now tangent is also an odd function. So the tangent of negative data is going to be equal to negative the tangent of data. Now looking at our expression here, the cosecant of negative pi over six earlier, we saw that the cosecant is an odd function. But if you forget that that's totally OK, because we know that the cosecant can be rewritten as a one over the sign. So this is just one over the sign of negative Pi over six. And based on my even odd identity here, the sign of negative pi over six is going to be equal to negative sign of positive pi over six. So I can simply find my sign value here using my left hand rule. Now the sign of pi over six is going to be equal to just one half. So this is really just equal to one over negative one half. Now simplifying this, this just ends up being negative two. So using our even odd identities to get there, we now know that the co sequent of negative pi over six is simply equal to negative two. Now how do we know when to use these identities? Well, as we continue to work through problems, we're going to have to use more and more identities. So how do we know exactly when to use our even odd identities specifically? Well, whenever your argument is negative, you know, to use your even odd identities, now, your argument is just whatever you're taking the cosine sign or tangent of. So if whatever you have in your parenthesis is negative, you know that you should use your even odd identities. Now that we've learned our even odd identities. Let's get some practice with them. Thanks for watching and I'll see you in the next one.

Example 1

We just saw how to use our even odd identities in order to rewrite expressions with negative arguments. And we specifically saw 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 trick function. So here in our first example, we have the negative tangent of negative data. That's this is already in terms of just one trick function, but we need to get rid of that negative argument there. So I know that the tangent of negative data is equal to negative tangent of data. So I can simply replace the tangent of negative data 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 data, I can replace that with negative tangent of data. 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 data. And now we have successfully rewritten this with no negative arguments in terms of just one trick function. So we're done with that example there, let's move on to our second example. Here we're asked to s to rewrite this expression, the sign of negative data over the cosine of negative data. Now, here I can go ahead and use my even odd identities for S and cosine. And I know that the sign of negative data is equal to negative sine of data, then the cosine of negative data because it's an even function is simply equal to the cosine of data. Now, from here, we have gotten rid of those negative arguments but it's still in terms of two trick functions. So how can we simplify this further? Well, the sign over the cosine is simply the tangent. So keeping that negative sign, this ends up just being negative tangent of theta. 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 an expression. So what you could have done here is recognized that the sign of negative data over the cosine of negative data is simply equal to tangent of negative data. Then using our even odd identity for tangent, we know that the tangent of negative data is simply equal to negative tangent of data. So we would end up getting that exact same answer, negative tangent of data. 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 six plus the cosine squared of 11 pi over six. 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 one? Well, here, I'm gonna show you exactly how to get there using what's called the Pythagorean identities, which we're going to use specifically to simplify tig expressions with squared trig functions. Now, here I'm gonna 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 one 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 links of Y and X and my hypotenuse of one, I end up with Y squared plus X squared is equal to one squared, which I know is just equal to one. Now, with this equation here, I can actually get even more specific with these side links because taking a deeper look at my triangle here with this angle of pi over six. And this hypotenuse of one 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 six. And my side length of Y is equal to the sine of pi over six. So plugging these values into my equation over here, I end up with the sine squared of pi over six plus the cosine squared of pi over six is simply equal to one. But you don't have to take my word for it that this is equal to one because we know these values right, the sine of pi over six is equal to one half. And the cosine of pi over six is equal to the square root of 3/2. So taking those values and squaring them, I end up with 1/4 plus 3/4. Now 1/4 plus 3/4 is equal to 4/4, which we know is just equal to one. So the sine squared to pi over six plus the cosine squared to pi over six is indeed equal to one. 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 one. Now, we can all derive our other two 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 one is equal to the CC squared of data. And then if I instead divided that first identity by the sign, I would end up with one plus the cotangent squared of theta is equal to the cosecant squared of the. And looking at my expression over here, the sine squared of 11 pi over six plus the cosine squared of 11 pi over six, seeing as these angles are the same just like I see in my identity here. I know that this is simply equal to one. 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 C squared of data minus the tangent squared of data. 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 C squared of the minus the tangent squared of data, 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 one is equal to the SN squared of theta plus one. Now these have the same two 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 one is equal to the SN squared of the. Now looking at this, knowing what my goal is here. If I were to go ahead and subtract the tangent squared of data from both sides, it would end up canceling on that left side just leaving me with one. And then on that right side, I'm left with the C squared of the minus the tangent squared of data. Now this is the exact expression that we were looking for. So that tells me that this is literally just equal to one. So now the C can squared of theta minus the tangent squared of data is simply equal to one using our Pythagorean identity in a different form. Now, let's take a look at another example here, here we have one minus the cosine of beta times one plus the cosine of data. And our goal here is still the same to rewrite this expression as a single term. Now here, it might not be obvious how we can use our Pythagorean identities because we don't even have a square trig function yet. But if I take these two terms and multiply them together noticing that this is one minus the cosine of the times one plus the cosine of data, I can go ahead and use my difference of squares formula to get that this multiplied out is one minus the cosine squared of data. So now we have a trig function that's squared and looking up at my Pythagorean identities here, I see the only identity I have that has the cosine squared is this first one, the sine squared plus the cosine squared is equal to one. So going ahead and rewriting that identity down here, the sine squared of theta plus the cosine squared of theta is equal to one. Now, this does not look like exactly what we're looking for. Yet. So how can we manipulate this in order to give us what we want, we want one minus the cosine squared of the. Now looking at this identity, if I subtract the cosine squared from both sides, it will cancel on that left side, leaving me with just the sine squared of theta. Then on the right side, I have one minus the cosine squared of theta, which is the exact expression that I'm looking for one minus the cosine squared of data. So now I see that one minus the cosine squared of data is simply equal to the sine squared of data. So that gives me my final answer that this is equal to sine squared of data. So we have successfully simplified this expression down to just one term. Now that we've seen how to use our Pythagorean identities regardless of what form they're in. Let's continue practicing. Thanks for watching and I'll see you in the next one.

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 1/1 plus the cosine of the. Now, it might not be immediately apparent how we can use our Pythagorean identities here because we don't even have a squared trick function. But let's play around with this a little bit in my denominator, I have one plus the cosine of the. Now, let's see what happens if I multiply this by one minus the cosine of data. Now remember we're working with a fraction. So if I'm multiplying the denominator by something, I need to multiply the numerator by the same exact thing. So as not to change the value of my expression, we're really just multiplying it by one. Now, let's see what happens when I multiply this out here in my numerator, I of course, just have one minus the cosine of theta having multiplied that by one. But in my nominator one plus the cosine of the times one minus the cosine of the, using my difference of squares, I get that this is one minus the cosine squared of data. Now I have a squared trick function and I can recognize this as being just a different form of that first Pythagorean identity. So if I subtract the cosine squared of data from both sides of this Pythagorean identity, it leaves me with the S squared of data is equal to one minus the cosine squared of data, 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 one plus sum trig function, if you multiply it by one minus that same trig function, you'll end up with a Pythagorean identity. Now using that here, I can replace that denominator with the sine squared of data. And in my numerator, I still just have that one minus the cosine squared of data. 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 one over the sine squared of theta times one minus the cosine of data. Now, the only reason that we have a fraction here is this one over sine squared. But remember one over the sine of data is just the C can. So one over the sine squared is simply the CO C squared of the. So now I'm just left with the CO C can squared of theta times one minus the cosine of the no more fraction. So this is my final answer here. That 1/1 plus the cosine of data is equal to the cos squared of the times one minus the cosine of data. 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 and Trig identities. And we will now be asked to simplify some more complicated trick 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 tig 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 trigger 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 data times the cos of data. Now, as we're simplifying these trigger expressions, it's important that we know what it means for a trigger expression to be fully simplified there. So there are three things that tell us a trigger expression is fully simplified. The first of those is that all of our arguments must be positive. Now, looking at my expression here, I see that I'm taking the tangent of negative data. So how can we fix that? Well, looking at our strategies here, our very first strategy and probably the most important one is that we want to be constantly scanning for identities. Now, looking at my identities here, since I have the tangent of negative data, that should tell me that I need to take a look at my even odd identities which tell me that the T of negative data is equal to the negative tangent of data. So using that identity here, I can rewrite this to have positive argument. So the tangent of negative data using that identity becomes the negative tangent of data. Now my expression is the negative tangent of data times the cos of data. And now all of my arguments are positive. Now let's take a look at the two other things that tell us a trigger 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. Now, looking at my expression here, I have two trig functions I have the tangent and I have the kit. So how can we get that to be fewer trick functions while taking a look at our strategies here, we are still constantly scanning for identities. And our next strategy tells us that we should add any fractions, but we don't have any fractions here. Then looking at that third strategy, my third strategy is going to be to break down everything in terms of cosine and sign. Now remember looking at this, I have two trick options, the tangent and the cot and taking a look at my identities here. I know that the cot is equal to one over sign and the tangent is equal to sign over cosine. So I can use those two identities in order to break this down in terms of just cosine and sign. So the negative tangent of data becomes negative sign of data over the cosine of data. Then the Cosan of data is just one over the sign of data. Now having broken this down in terms of sine and cosine allows me to cancel some stuff out because I have sign on the top and sign on the bottom that goes away and I am just left with negative one over the cosine of the. So now I have just one trick 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 here. I know that one over the cosine is the C A. So this negative one over cosine that I have in my expression can be simplified down to the negative C of data. 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 data times the cos of the fully simplifies down to the negative C of data. Now, you may have noticed here that we didn't use all of our strategies and that's totally OK. 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. Now, 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 one plus the cosine of the. Now remember for this trigger 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 trick 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 here, remember we are constantly scanning for identities. We want to add any fractions which we can't really do that here. And 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 gonna help me. Now taking a look at our fourth strategy here, if we have one plus or minus sum trig function, we want to multiply both the top and the bottom of our fraction by one minus or plus that same trig function. Now looking at my expression, I have one plus the cosine of data. So using that strategy, I multiply this by one minus the cosine of data 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 new reader, I'm going to leave that factored as the sine squared of theta times one minus the cosine of the. Now in my denominator, I recognize this as being a difference of squares because I have the same two terms one and cosine just one with a plus and one with a minus. So multiplying this out gives me one minus the cosine squared of data. 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 data plus the cosine squared of data is equal to one. And if I subtract cosine squared of data from both sides here, it will cancel on that left side leaving me with exactly what's in the denominator of my expression here, one minus the cosine squared of the. Now I know that one minus the cosine squared of data is equal to the sine squared of data. So I can replace my denominator. Now, in my numerator, I'm keeping that the same for now the sine squared of data times one minus the cosine of data. But in my denominator, I am using that Pythagorean identity in order to replace that with the sine squared of data. Now, I have the sine squared in both the numerator and the denominator and that cancels out. Now all I'm left with is one minus the cosine of the. So let's figure out if we need to do anything else here. Here we see the, 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. So we have now successfully fully simplified this down from the sine squared of the over one plus cosine of data down to just one minus the cosine of data. 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 trigger expressions, let's continue practicing together. Thanks for watching and I'll see you in the next one.

Example 3

Everyone in this problem, we're asked to simplify the expression and the expression we're given is the sine squared of the minus the tangent squared of data over the sine of data plus the tangent of data. 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 S squared of data minus the tangent squared of data. You may recognize this as being a difference of squared because this is one term squared minus another term squared. So this will factor to that first term pull that second term times that first term minus that second term. So factoring that out I can get the sign of data that first term plus the tangent of data. Second term and then multiplying that by the sign of data minus the tangent of data. Now my denominator stays the same here, the sign of the plus the tangent of data. But here I noticed that I can cancel some 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 sign of data minus the tangent of data. 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 two trick functions but because they're being subtracted, it's gonna be kind of hard to make it so that there are any less trig functions that are already there. So having these two 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 tig expression down to the sign of data minus the tangent of data. 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 data plus the cosecant of data divided by the cosine of data plus the sine of data minus the sequence of the over the sine of data. 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 are 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 S. 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 sign. 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 data over the sine of data. And then I'm going to multiply my second fraction by the cosine of data over the cosine of data. 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 sign into these two terms here. Starting off my denominator with sign of data times cosine of data plus the sign of the times the cosecant of data. 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 the times the S of the and then minus because that's a subtraction right there minus the cosine of the times the s of data. OK. 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 S and cosine. Now looking down to my identities here, I know that the CO sn is equal to one over sine and the SN is equal to one over cosine. So I can break this entire thing down in terms of sine and cosine by using those identities to replace the coin and the sequin in this expression. So let's go ahead and do that here. Now everything is staying the same besides the cosecant and the SN. So I still have that first term sign of the times cosine of data. But now my second term is sign of the times one over the sign of data using that cos and reciprocal identity. Then that other term will say the same cosine of the times the sine of data. And then this term becomes cosine of the times one over the cosine of data. Now, remember we still have a fraction, we still have our denominator here and it is still the cosine of the times the sine of data. Now, the only thing that I changed was this to one over S sign and this to one over cosine. But let's look at what happens here. Now, here I have a sign on the top and I have it being multiplied by one over the sign. So those are going to cancel. Now, when those cancel out, that's just going to leave me with a value of positive one. 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 one and subtract one, I get zero? So those fully go away canceling each other out. Now, all that I'm left with, in that numerator is S the 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 the times the sine of data. 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 data times the cosine of data plus the cosine of data times the sign of data. 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 two. So this is really just two times the sine of data times the cosine of data in that numerator. Now this is still being of course divided by the cosine of data times the sign of data. But again, let's look at what's really happening here. I have a sign on the top. I have a sign on the bottom. I have a cosign on the top and I have a cosign on the bottom. So what's the only thing that I'm left with two? And that's my final answer here because I don't have any fractions. And I have as few trick functions as possible because I have literally no trick functions. So our original expression here, this long mess of fractions is really just equal to two. 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

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. Now, 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 to walk you through exactly how to do that here. So let's go ahead and get started. Now in working with these problems, you may only have to simplify one side of your equation or you may have to simplify both sides. And this is something that will become more apparent as you begin to work through a problem. So let's go ahead and jump right into our first example here where we're asked to of course, verify the identity. Now, 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. Now the identity that we're asked to verify here is sine theta times cosine theta over one minus the cosine squared of data. And this is equal to one over the tangent of data. 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. Now, looking at my left side sine theta times cosine theta over one minus the cosine squared of data. Remember one of our most important strategies for simplifying is to constantly be scanning for identities and looking at this left side of my equation. This one minus cosine squared of data looks familiar and looking at my identities here, I know that one of my Pythagorean identities tells me that the sine squared of data plus the cosine squared of data is equal to one. So if I subtract the cosine squared of data from both sides of this identity, I end up with exactly what's in that denominator on that right side, one minus the cosine squared of data. So using this identity, I can replace that denominator with the sine squared of data using that Pythagorean identity. Now I'm going to keep my numerator the same sine the times cosine data and I can do some canceling here because in my denominator, I have sine squared of data. And in my numerator, I have the sign of data so that s data in my numerator goes away. And in my denominator, my sign is no longer squared and I'm simply left with the cosine of data over the sine of data. Now remember we are still constantly scanning for identities here and I know that cosine data over sine of data is simply the cotangent of data. Now, I also know that the cotangent is equal to one over the tangent, which is exactly what the right side of my equation is. And remember that our old goal is to show that these two sides of our equation are equal to each other. So knowing that this cosine over sign is one over the tangent, I can go ahead and rewrite that side as one over the tangent of data. Now, I don't have to do anything to that right side of my equation. It's just one over the tangent of the. And now we have that one over the tangent is equal to one over the tangent. 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 that you have to learn all that we're doing here is showing that two sides of an equation are equal. So let's go ahead and take a look at our second example. Here, we're still verifying the identity. But the trigger equation that we're given here is the second squared of data minus the tangent squared of data over the cosine of negative data plus one is equal to one minus the cosine of the over the sine squared of data. 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 and that's totally OK. Now, 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, the C squared of theta minus the tangent squared of data over the cosine of negative data plus one. Now, remember here, we want to be constantly scanning for identities and looking at this particular side of my equation, this C can squared minus the tangent squared reminds me of a Pythagorean identity which tells us that the tangent squared of theta plus one is equal to the C squared of data. So if I rearrange this identity and subtract the tangent squared of data from both sides, I will end up with exactly what's in that numerator there. And it's simply equal to one. So I can replace the entire entire numerator with just one. Then in my denominator, I still have that cosign of negative theta plus one. Now remember we're still constantly scanning for identities here and looking in that denominator since I have the cosine of negative data, 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 data is simply the cosine of data. So simplifying that denominator further leaving my numerator as one because that cannot be further simplified in my denom. I now just have the cosine of the plus one. 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 one minus the cosine of data over the S squared of data. Now, one of my simplifying strategies tells me that if I have one plus or minus a trig function, I wanna go ahead and multiply the top and the bottom by one minus or plus that same trig function. Now here in my numerator, since I have one minus cosine of the, I'm going to go ahead and multiply this by one plus the cosine of the. And remember if I'm multiplying the top by that, I need to multiply the bottom by that as well as not to 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 one minus cosine squared of theta. Then in my denominator, I'm going to keep that as is sine squared of theta times one plus the cosine of data. Now here in my numerator scanning for identities, this one minus cosine squared of data is an identity that we've used before our pythagorean identity. Now I know that one minus the cosine squared of data is simply equal to the sine squared of data. So I can go ahead and rewrite that numerator as the sine squared of data. And then in my denominator, I also have a sine squared of data and that is still multiplied by one plus the cosine of data. Now, here I can cancel some stuff out because I have S squared on the top and S squared on the bottom. Now I am simply left with a one in that numerator and then a one plus the cosine of theta 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 one and the cosine to get them exactly the same one plus the cosine of data. 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 the one minus sine of the over the cosine of the minus cosine of the over one plus the sine of data is equal to zero. So it's very clear what side we should be working with this left side because we can't do anything to that zero 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 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 cosine of data and one plus the sine of data. So my common denominator will take both of these and multiply them together to give me a new denominator of the cosine of data times one plus the S of data. Now, in order to get at 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 one plus the sign of data. And if I'm multiplying the bottom by that, I also need to multiply the top by that one plus the sine of data. Now multiplying that out will end up giving me a numerator of one minus the S squared of data because this is a difference of squares. Now let's look at that second fraction. We want to multiply both of these by the cosine of data in order to get that common denominator. Now, on the top, I am left with that minus cosine squared of data, 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 one minus the science squared of data and coming over here to my identities, I know that my first and identity tells me that the sine squared plus the cosine squared is equal to one. So if I subtract the sine squared of theta from both sides, it will cancel on that left side and leave me with that one minus the sine squared of theta, which is exactly what I have highlighted here in my numerator. So that one minus sine squared is equal to the cosine squared of data. So now in my numerator using that Pythagorean identity, I am left with the cosine squared of data minus the cosine squared of theta. And then all of that is over that same denominator, the cosine of the times one plus the sine of the. Now what's happening here? Well, in my numerator, I have the cosine squared minus the cosine squared. So what happens when I take something? And I subtract that same something? Well, I'm just left with zero. Now, that right side of my equation is also zero. So I have successfully verified this identity because zero is equal to zero. 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. And the identity that we're given here is the sin of the times one minus the sine squared of the is equal to one plus the tangent squared of the times the cotangent squared of data divided by the cos and squared of the times the CN of data. 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, one plus the tangent squared times the code tangent squared over the code C squared times the second squared. Where can we go from here? Well, remember we want to be constantly scanning for identities. And the first thing that I noticed here in my numerator is one plus the tangent squared. Now one plus the tangent squared, taking a look at my Pythagorean identities is equal to the C squared. So I can go ahead and replace that in my numerator there. So I have the C squared of the times the co tangent squared of data with that same denominator, the cos N squared of data times the CN of data. Now, here I can go ahead and just cancel this see on the bottom with one of the SN on the top. So now I am left with the CN of the times the cotangent squared of data over the cos N squared of data. Now, where can we go from here? Well, remember, let's take a look at our strategies here and 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 one plus or minus a trick function. And we can't factor here either. So let's go ahead and break this down. Now, I know that the C of data is equal to one over the cosine of data. And the co tangent squared of data is equal to the cosine squared of data over the sine squared of data. Then in my denominator, I can also rewrite that in terms of sign it is one over the sine squared of data. 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 data over the sine squared of data. 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 the over one. Now, here I have sine squared on the top S squared on the bottom that goes away. And all I'm left with is the cosine of data. 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 data. Now looking at my left side, the CN of theta times one 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 one minus the sine squared of data, which I know is one of my pythagorean identities just in a slightly different form because the sine squared of data plus the cosine squared of data is equal to one. So if I go ahead and subtract the sine squared of data from both sides, I am left with exactly what's in those parentheses there. And that will be equal to the cosine squared of data. So I can rewrite this as the sequence of the times cosine squared of data. Now, where can we go from here, well, the sequence of the I know is one over the cosine of data. And if this is multiplying the cosine squared of data, that's gonna be really helpful. So what can we do here? We have a cosine on the 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 data. 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 data is equal to the cosine of data. 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.