Trigonometric Identities - Online Tutor, Practice Problems & Exam Prep

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 θ times the cosecant of θ. Now, 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 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 θ. 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 θ, that should tell me that I need to take a look at my even-odd identities, which tell me that the tangent of negative θ is equal to the negative tangent of θ. So using that identity here, I can rewrite this to have a positive argument.

So the tangent of negative θ using that identity becomes the negative tangent of θ. Now, my expression is the negative tangent of θ times the cosecant of θ. And now, all of my arguments are positive. Now let's take a look at the 2 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. Now looking at my expression here, I have 2 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 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 sine. Now remember, looking at this, I have 2 trig functions, the tangent and the cosecant. And taking a look at my identities here, 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 2 identities in order to break this down in terms of just cosine and sine.

So the negative tangent of θ becomes negative sine of θ over the cosine of θ. Then the cosecant of θ is just one over the sine of θ. Now 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 θ.

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 here, 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 θ.

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 θ times the cosecant of θ fully simplifies down to the negative secant of θ. 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. Now in this next example, we are still simplifying the expression, but the expression that we're given here is the sine squared of θ over 1 plus the cosine of θ. 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 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 going to help me.

Now taking a look at our 4th 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 θ. So using that strategy, I want to multiply this by 1 minus the cosine of θ, 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 θ times 1 minus the cosine of θ.

Now in my denominator, I recognize this as being a difference of squares because I have the same two terms, 1 and cosine, just 1 with a plus and 1 with a minus. So multiplying this out gives me 1 minus the cosine squared of θ. 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 θ plus the cosine squared of θ is equal to 1. And if I subtract the cosine squared of θ from both sides here, it will cancel on that left side, leaving me with exactly what's in the denominator of my expression here, 1 minus the cosine squared of θ.

Now I know that 1 minus the cosine squared of θ is equal to the sine squared of θ. So I can replace my denominator. Now in my numerator, I'm keeping that the same for now, the sine squared of θ times 1 minus the cosine of θ. But in my denominator, I am using that Pythagorean identity in order to replace that with the sine squared of θ. 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 θ. 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.

So we have now successfully fully simplified this down from the sine squared of θ over 1 plus the cosine of θ down to just 1 minus the cosine of θ. 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.

Simplifying Trig Expressions Example 1

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, 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 we do have one big fraction here, 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. 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 into that first term plus that second term times that first term minus that second term. Factoring that out, I can get the sine of theta, that first term, plus the tangent of theta, 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 is now 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 trigonometric functions as possible. Now here I have 2 trigonometric functions.

However, because they're being subtracted, it's going to be kind of hard to make it so that there are any fewer trigonometrical functions than there already are. So having these 2 trigonometrical functions is as few trigonometrical functions as I can have possibly. We have also satisfied that last piece of criteria there, and we have fully simplified our trigonometric expression down to the sine of theta minus the tangent of theta. Thanks for watching, and let me know if you have any questions.

sin⁡θ2 - tan⁡θ2 sin⁡θ + tan⁡θ

Simplifying Trig Expressions Example 2

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 trigonometric 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 at the bottom is going to give me that common denominator there. But we need to multiply 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 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 1 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 one over the cosine of theta. Now remember, we still have a fraction. We still have our 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 trigonometric expressions, you're going to come across a new type of problem in which you're given a trigonometric 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 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. 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 sinθ∙cosθ1-cosθ2 and this is equal to 1tanθ. 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 the right side. So we're going to start by applying our simplifying strategies to that left side.

Looking at my left side, sinθ∙cosθ1-cosθ2, remember, one of our most important strategies for simplifying is to constantly be scanning for identities. This 1-cosθ2 looks familiar. Using the Pythagorean identity sinθ2+cosθ2=1, we can replace the denominator with sinθ2.

Keeping the numerator the same as sinθ∙cosθ, some canceling allows us to have cosθsinθ in the numerator. Recognizing that cosθsinθ is the cotangent of theta, which is equivalent to 1tanθ on the right side, we conclude that both sides are equal.

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. It's all about 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 trigonometric equation that we're given here is secθ2-tanθ2cos(-θ)+1 is equal to 1-cosθsinθ2.

Start with the more complicated side, simplify using the identity tanθ2+1=secθ2 to replace all in numerator with 1. Use the even-odd identity to simplify cos(-θ) to cos(θ). Further mathematical manipulations give both sides simplifying to fa...

Verifying Trig Equations as Identities Example 3

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 1−sin⁡θ⁢cos⁡θ−cos⁡θ1+sin⁡θ is equal to 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 cosine of theta and 1 plus the sine of theta. So my common denominator will take both of these and multiply them together to give me a new denominator of the cosine of theta times 1 plus the sine of theta. 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 plus the sine of theta. And if I multiply the bottom by that, I also need to multiply the top by that, 1 plus the sine of theta. Now multiplying that out will end up giving me a numerator of 1 minus the sine squared of theta because this is a difference of squares. Now let's look at that second fraction. We want to multiply them both by the cosine of theta in order to get that common denominator.

Now on the top, I am left with that minus cosine squared of theta. Remember, 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 sine squared of theta.

Coming over here to my identities, I know that my first Pythagorean identity tells me that the sine squared plus the cosine squared is equal to 1. So if I subtract the sine squared of theta from both sides, it will cancel on the left side and leave me with that 1 minus the sine squared of theta, which is exactly what I have highlighted here in my numerator. So that 1 minus sine squared is equal to the cosine squared of theta. So now, in my numerator, using that Pythagorean identity, I am left with the cosine squared of theta minus the cosine squared of theta. And then all of that is over that same denominator, the cosine of theta times 1 plus the sine of theta.

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 subtract that same something? Well, I'm left with 0. Now that right side of my equation is also 0, so I have successfully verified this identity because 0 equals 0, and we're done here.

Let me know if you have any questions, and thanks for watching.

Verifying Trig Equations as Identities Example 4

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 secant of theta times 1 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 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 secθ2. 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 that same denominator, the cosecant squared of theta times the secant of theta. Now 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 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 1 plus or minus a trigonometric 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 1 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 going to 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 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 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.

Solve Trig Equations Using Identity Substitutions

Hey, everyone. When solving linear trig equations, we wanted to find an angle theta that made our equation true. And now that we're tasked with solving some more complicated trig equations, we still want to do the same thing. We want to find an angle theta that makes our equation true. But now that these equations have more than just one trig function, it's not quite so simple because how are we going to find an angle theta for which the secant squared of it minus 1 over the tangent of it is equal to 1?

I don't know how to do that just using the unit circle. But now that we know a bunch of different trig identities, we can now use those identities to rewrite these equations in terms of just one trig function. Then we're just left with a linear trig equation that we already know how to solve. So here I'm gonna walk you through exactly how to do that. Let's go ahead and get started.

Now here we have the equation, the secant squared of theta minus 1 over the tangent of theta is equal to 1. Now seeing that I have multiple trig functions here tells me that I need to go ahead and use my trig identities to rewrite this in terms of just one trig function. Now we're going to keep using our strategies for simplifying here. And scanning this for identities, I see in my numerator this secant squared of theta minus 1. Now whenever I have a squared trig function, I should be thinking about using my Pythagorean identities here.

Now seeing that I have this secant squared of theta minus 1, I can use my Pythagorean identity to rewrite this, giving that this is just the tangent squared of theta over the tangent of theta is equal to 1, having used that Pythagorean identity. Now from here, I can cancel some stuff because that tangent on the bottom will cancel with one of my tangents on the top, leaving me with just the tangent of theta is equal to 1. So we started with multiple trig functions and then used identities to get down to just one trig function. And from here, we can just find our angles theta on the unit circle, for which this is true, and then add this factor of pi n to account for all possible solutions. Now not all of these equations are going to be quite so simple, and we're not always just going to use our Pythagorean identities.

So let's go ahead and walk through another example together. Here we're asked to find all solutions to the equation sine of 2 theta over the cosine of negative theta is equal to 1. Now here I want to be constantly scanning for identities. And something that you might notice here is that we have a sine over a cosine. So your first instinct might be to say, okay, wouldn't that just be the tangent?

But we can't actually do that here because these arguments are not the same. In my numerator, I have 2 theta in my argument. And in my denominator, I have the negative theta. So I can't use that identity. I need to think of another way to simplify this.

Well, looking at that numerator, I have the sine of 2 theta. So having 2 theta in that argument, what identity should we be using? Well, here we can rewrite the top using a double angle identity since we have that 2 theta in the argument. So using that identity here, I end up with 2 times the sine of theta times the cosine of theta over the cosine of negative theta, keeping that the same for now, is equal to 1, having rewritten the top there. Now looking at this, I see in my denominator I have this negative theta.

So how do I get rid of that negative theta? Because I don't want a negative argument here. Well, whenever we have a negative argument, we want to think about using our even-odd identities. So here I can rewrite the bottom of this fraction using an even-odd identity. Now I know that the cosine is an even function, so I can rewrite that denominator as the cosine of theta.

Now everything else is staying the same for now. 2 times the sine of theta times the cosine of theta and all of this is equal to 1. Now what's going to happen here? Well, I see I have a cosine in the top, and a cosine in the bottom, so those just cancel out. Now all I'm left with is 2 times the sine of theta is equal to 1.

So I have just the sine of theta is equal to 1. So I have just one trig function now and now all that's left to do is solve this resulting linear trig equation. So let's go ahead and do that. Now I want to go ahead and isolate this trig function, so I'm going to divide both sides by 2. Now they'll cancel on that left side, leaving me with the sine of theta is equal to 1 half.

Now from here, I just want to think of my angles theta for which the sine is equal to 1 half on my unit circle. Now the angles for which that are true are theta is equal to pi over 6 and theta equals 5 pi over 6. Those are my two angles on the unit circle for which the sine is equal to 1 half. But remember that this problem asked us to find all solutions, so we need to go ahead and add 2 pi n to each of these. Now this represents all solutions to my original equation, and this is my final answer here.

Theta = π 6 + 2 π n and theta = 5 π 6 + 2 π n

Now that we know how to use identities to solve more complicated trig equations, let's continue practicing together. Thanks for watching and I'll see you in the next one.

Find all solutions to the equation.

$\left(\cos\theta+\sin\theta\right)\left(\cos\theta-\sin\theta\right)=-\frac12$

• A. $\theta=\frac{5\pi}{12}+2\pi n,\frac{7\pi}{12}+2\pi n$
• B. $\theta=\frac{2\pi}{3}+2\pi n,\frac{4\pi}{3}+2\pi n$
• C. $\theta=\frac{\pi}{3}+2\pi n,\frac{2\pi}{3}+2\pi n$
• D. $\theta=\frac{\pi}{3}+\pi n,\frac{2\pi}{3}+\pi n$

Find all solutions to the equation where 0 ≤ $\theta$ ≤ $2\pi$.

$\sin\theta\cos\left(2\theta\right)-\sin\left(2\theta\right)\cos\theta=\frac{\sqrt2}{2}$

• A. $\theta=\frac{5\pi}{4},\frac{7\pi}{4}$
• B. $\theta=\frac{\pi}{4},\frac{3\pi}{4}$
• C. $\theta=\frac{5\pi}{4}+2\pi n,\frac{7\pi}{4}+2\pi n$
• D. $\theta=\frac{3\pi}{4},\frac{7\pi}{4}$