Trigonometric Identities: Videos & Practice Problems

Understanding the properties of trigonometric functions is essential for simplifying expressions. The cosine function is even, meaning $f(-x)=f\left(x\right)$, while the sine function is odd, so $f(-x)=-f\left(x\right)$. Pythagorean identities, such as $\sin ²\left(\theta \right)+\cos ²\left(\theta \right)=1$, help simplify squared functions. Additionally, sum and difference identities allow for the evaluation of trigonometric functions with combined angles, enhancing problem-solving skills in trigonometry.

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 trigonometric 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 trigonometric problems.

So here, I'm going to break down for you exactly which trigonometric 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 π/2. Now here on my graph I see that f(π/2) is equal to zero and if I go to the other side of my graph and check out the value of my function for -π/2 I see that f(-π/2) is also equal to zero. So relating these two points to each other, I could say that f(-π/2) is equal to f(π/2) 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 -π/2, 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 π/2, I see that f(π/2) is equal to positive one. Then going to the other side of my graph looking at f(-π/2), I see that f(-π/2) is actually equal to negative one. So relating these two points to each other, I could say that f(-π/2) is equal to -f(π/2) because my value for f(-π/2) was negative one, and for positive π/2 was positive one.

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 flipped the sign of our input value to -π/2, 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 trigonometric functions? Well, we already know that secant is one over cosine. So with that in mind, it makes sense that secant 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 sine. But what about tangent? Well, we know that tangent is sine over 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 trigonometric 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 expressed 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 x, y, a, b, 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 theta, knowing that our cosine is an even function, we know that the cosine of negative theta is simply equal to positive cosine of theta. So here with our expression, the cosine of -π/4, I know that this is really just equal to the cosine of π/4. 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 two over two. So we didn't have to figure out where -π/4 is on our unit circle because we have our even odd identities. Now taking a look at sine of negative theta here, we know that sine is an odd function.

So the sine of negative theta is simply equal to negative sine of theta. Now tangent is also an odd function. So the tangent of negative theta is going to be equal to negative the tangent of theta. Now looking at our expression here, the cosecant of -π/6 earlier, we saw that the cosecant is an odd function but if you forget that that's totally okay because we know that the cosecant can be rewritten as one over the sine so this is just one over the sine of -π/6. And based on my even odd identity here, the sine of -π/6 is going to be equal to negative sine of positive π/6.

So I can simply find my sine value here using my left-hand rule. Now the sine of π/6 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 cosecant of -π/6 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, sine, 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.

Even and Odd Identities 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 trig function. So here in our first example, we have the negative tangent of negative theta. Now this is already in terms of just one trig function, but we need to get rid of that negative argument there.

So 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 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 trig 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 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 recognize that the sine of negative theta over the cosine of negative theta is simply equal to tangent of negative theta. 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. $-2\cos\theta$

Answer: (A)

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. $\frac{\cos\theta}{\sin^2\theta}$

Answer: (C)

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

Answer: (C)

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. $\frac{1}{-\sin38\degree}$

Answer: (B)

Pythagorean Identities

Hey everyone. As you continue to work through problems dealing with trigonometric expressions, you may come across an expression that looks like this one: sin211π6 plus cos211π6. Now, if this is the first time that you are seeing an expression with multiple squared trigonometric functions, this can look really intimidating. But what if I told you that this was just equal to one? Well, 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 trigonometric expressions with squared trigonometric functions.

Now, 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 one without a second thought. So let's go ahead and get started. 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 a2+b2=c2.

If I plug in my side lengths of y and x and my hypotenuse of one, I end up with y2+x2=1, which I know is just equal to one. 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 π6 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 π6 and my side length of y is equal to the sine of π6. So plugging these values into my equation over here I end up with sin2π6 plus cos2π6 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 π6 is equal to one half and the cosine of π6 is equal to the square root of three over two. So taking those values and squaring them, I end up with one fourth plus three fourths. Now one fourth plus three fourths is equal to four over four, which we know is just equal to one. So sin2π6 plus cos2π6 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 sin2θ+cos2θ=1. Now we can also 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 tan2θ+1=sec2θ. And then if I instead divided that first identity by the sine I would end up with 1+cot2θ=csc2θ. And looking at my expression over here sin211π+cos211π, 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 trigonometric 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 trigonometric 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 sec2θ-tan2θ and here we're asked to use our Pythagorean identities to rewrite the expression as a single term. Now looking at this expression here, the sec2θ-tan2θ, 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, tan2θ+1=sec2θ.

Now these have the same two trigonometric functions that I see in my example here. So let's go ahead and see what we can do with this trigonometric identity. tan2θ+1=sec2θ. Now looking at this, knowing what my goal is here, if I were to go ahead and subtract the tan2θ from both sides, it would end up canceling on that left side, just leaving me with one, and then on that right side, I am left with sec2θ-tan2θ. 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 sec2θ-tan2θ 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 θ times one plus the cosine of θ, 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 squared trigonometric function yet. But if I take these two terms and multiply them together, noticing that this is one minus the cosine of θ times one plus the cosine of θ, I can go ahead and use my difference of squares formula to get that this multiplied out is one minus the cosine squared of θ.

So now we have a trigonometric 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. sin2θ+cos2θ=1. So going ahead and rewriting that identity down here, sin2θ+cos2θ=1. 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 θ. 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 θ. Then on the right side, I have one minus the cosine squared of θ, which is the exact expression that I'm looking for one minus the cosine squared of θ. So now I see that one minus the cosine squared of θ is simply equal to the sine squared of θ, so that gives me my final answer, that this is equal to sine squared of θ.

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.

Pythagorean Identities Example 2

Hey everyone! In this example problem, we're asked to use our Pythagorean identities to rewrite the expression with no fraction, and the expression that we're given here is 11+cosθ. 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+cosθ. Now, let's see what happens if I multiply this by 1-cosθ.

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 1-cosθ having multiplied that by one, but in my denominator 1+cosθ times 1-cosθ using my difference of squares I get that this is 1-cos2θ. 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 cos2θ from both sides of this Pythagorean identity, it leaves me with sin2θ is equal to 1-cos2θ, 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 some 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 sin2θ and in my numerator, I still just have that 1-cosθ. 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 1sin2θ times 1-cosθ. And now the only reason that we have a fraction here is this 1sin2θ. But remember, one over the sine of theta is just the cosecant.

So one over the sine squared is simply the cosecant squared of theta. So now I'm just left with the cosecant squared of theta times one minus the cosine of theta. No more fraction. So this is my final answer here that 11+cosθ is equal to the cosecant squared of theta times one minus the cosine of 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. $-\cot^2\theta$

Answer: (D)

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. $-\tan^2\theta\left(1+\sec\theta\right)$

Answer: (C)

Sum and Difference of Sine & Cosine

Hey, everyone. We just learned a bunch of different trig identities that we've used to simplify trig expressions, but we're not quite done yet. Looking at this expression here, the sine of π/2 plus π/6, I can't simplify this using any of the identities that we already know, so we need something new. And that something new is our sum and difference identities. Now here I'm going to walk you through exactly what our sum and difference identities are and how we can use them to continue simplifying trig expressions and also to find the exact value of trig functions without using a calculator even if they're not on the unit circle.

So let's go ahead and get started. Now as you might have noticed from this expression here, our sum and difference identities are going to be useful whenever we have multiple angles in the argument of a trig function. Like here we have the sine of π/2 plus π/6, two separate angles inside that single argument. Now we can use our sum and difference identities in order to expand this out and make it easier to work with. So looking at our first identity here, the sine of some angle a plus another angle b, is equal to the sine of that first angle a times the cosine of that second angle b plus the cosine of that first angle a times the sine of that second angle b.

So looking again at our expression here, the sine of π/2 plus π/6, which you might also see written in degrees as the sine of 90 degrees plus 30 degrees, we can use this formula here in order to expand this out. So let's go ahead and do that. Now using that formula this gives me the sine of that first angle π/2 times the cosine of that second angle π/6 plus the cosine of that first angle, again π/2, times the sine of π/6. Now from our knowledge of the unit circle, we can break this down even further because we know that the cosine of π/2 is just zero, so this entire term will go away. Then we also know that the sine of π/2 is equal to one, so this will leave me with just the cosine of π/6, which is much easier to deal with than the sine of π/2 plus π/6.

Now again, from our knowledge of the unit circle, we know that the cosine of π/6 is equal to the square root of three over two, and we're done here. Now looking at our original expression, you may have been tempted here to just add these two angles together and evaluate the sine of that directly, which is something that can work sometimes, but it's not always going to make working with these expressions easier. So here we saw that this simplified down to the cosine of π/6, which is much easier to work with. Now let's take a look at our other sum and difference identities here. The sine of some angle a minus an angle b is almost identical to our formula for the sine of a plus b, except now we're just subtracting those two terms.

Now we also want to look at our cosine identities here. And here the cosine of an angle a plus another angle b is going to be equal to the cosine of a times the cosine of b minus the sine of a times the sine of b. Now here you'll notice that here we're subtracting these two terms even though our two angles are being added together. So it's kind of backward to what you might think it is. Now our difference formula for cosine is again almost identical, but here we're actually going to be adding those two terms together.

Now that we know all of our sum and difference identities, how do we know when to use them? Well, something that might be kind of obvious is that whenever our argument contains a plus or a minus, we should use our sum and difference identities. Like here, we saw the sine of π/2 plus π/6. Now it all isn't always going to be that obvious that we should use our identities. So we should also use them whenever we have an argument with multiples of either 15 degrees or π/12 radians.

Now how do we recognize that? Because that seems like kind of an obscure piece of criteria that tells us to use our sum and difference identities. So let's dive a little bit deeper into this. Now what if I asked you to find the exact value of the function the cosine of 15 degrees without using a calculator? That might be kind of tricky because 15 degrees is not an angle that's directly on the unit circle that we already know everything about.

So let's instead think of angles that we do know from the unit circle. We know 60 degrees, 45 degrees, 30 degrees, these are all angles that we're really familiar with. So what if instead of taking the cosine of 15 degrees, I took the cosine of some combination of these angles that I do know. Now I could try to take the 60 degrees minus 30 degrees, but that's just 30 degrees. But I could take 45 degrees and subtract 30 degrees, and that's equal to 15 degrees.

So what if instead here I took the cosine of 45 degrees minus 30 degrees? Then couldn't I find the exact value of this original function using angles that I already know? Well, that's exactly what would happen. So whenever we're faced with finding the exact values of trig functions for angles that are not on the unit circle, we simply can rewrite our argument as the sum or difference of two known angles. Then we can just use our sum and difference identities in order to expand this out and use a bunch of trig values that we already know.

So let's go ahead and do that here. Here we have the cosine of 45 degrees minus 30 degrees. So I can go ahead and use my difference formula for cosine to expand this out. Cosine of 45 degrees minus 30 degrees will give me the cosine of that first angle 45 degrees times the cosine of that second angle 30 degrees. Then I add that together with the sine of 45 degrees, that first angle again times the sine of 30 degrees.

Now these are things that I already know from the unit circle. I know all of these trig values. So replacing all of these, substituting those values in, the cosine of 45 degrees is equal to the square root of two over two.

The cosine of 30 degrees is equal to the square root of three over two, and then I'm adding that together with the sine of 45 degrees, which is again the square root of two over two, times the sine of 30 degrees, which is simply one half. Now from here, I can multiply these fractions across just doing some algebra to get to my final answer. Now multiplying these fractions together, the square root of two over two times the square root of three over two gives me the square root of six over four. Then I'm adding that together with these two fractions multiplied by each other, which will give me the square root of two over four. Now these both already have a common denominator, so I can simply rewrite this as the square root of six plus the square root of two all over four, and that is my final answer.

So now that we know how to use our sum and difference identities, let's continue practicing with them. Thanks for watching, and I'll see you in the next one.

Sum and Difference of Sine & Cosine Example 3

Hey everyone, in this problem we're asked to rewrite each argument as the sum or difference of two angles on the unit circle. Now this would be helpful if you were asked to find the exact value of an angle, like say the sine of 75 degrees. You probably don't know the trigonometric values of 75 degrees off the top of your head. If instead, we rewrite this in terms of angles that we do know on the unit circle, we can then use a sum or difference formula to get to an answer much quicker. So let's look at this first example.

Here we have 75 degrees, and we want to rewrite that as a sum or difference of angles on our unit circle. I'm going to focus on angles that are in the first quadrant of the unit circle because those are the angles that I'm most familiar with. I'm going to look at 30 degrees, 45 degrees, and 60 degrees, and I want to find some combination of two of these angles that add or subtract to 75 degrees. Looking at these angles the first thing that I notice is that if I have 30 degrees and I add that together with 45 degrees, that gives me the 75 degrees that I'm looking for. If I took the sine of 30 degrees plus 45 degrees and then used a sum formula, I could get to my answer.

Now let's look at a second example. Here we have the cosine of negative 15 degrees. Again, we want to find two out of these three angles that either add or subtract to that negative 15 degrees. Looking at these, I have my 45 degrees and my 30 degrees, and if I subtracted 45 from 30, that would give me negative 15. So here, if I took 30 degrees minus 45 degrees and I took the cosine of that and then used a difference formula, I could then get to an answer.

Now let's look at one final example. Here we have the cosine of seven pi over 12. Seven pi over 12 is in radians, so this can be a little trickier to work with because it's a fraction. But remember that our angles are still these three angles. We have pi over six, pi over four, and pi over three.

We want to find two of these angles that add or subtract to give us seven pi over 12. Looking at seven pi over 12, I want to think of numbers that either add or subtract to that. So, if I think of that seven and I think of two numbers that add or subtract to seven, I first think of three and four. Now here, if I took three pi over 12 and I added it together with four pi over 12, that gives me seven pi over 12.

But these don't look quite like these angles yet, but let's simplify these. Three pi over 12 can be simplified to pi over four and four pi over 12 can be simplified to pi over three, which are two of my three angles here. So here I could take the cosine of pi over four plus pi over three then use my sum formula to get an answer. Now earlier we saw that we want to use our sum and difference identities whenever we see angles that are multiples of 15 degrees or pi over 12 radians and that's exactly what we see here. Here we have the sine of 75 degrees.

75 is a multiple of 15, as is negative 15. Seven pi over 12 is a multiple of pi over 12. So whenever we see that, we want to go ahead and break down our argument as a sum or difference and then use our identities from there. Thanks for watching, and let me know if you have questions.

Find the exact value of the expression.
$\cos105\degree$

• A. $\frac{\sqrt2-\sqrt6}{4}$
• B. $\frac{\sqrt6-\sqrt2}{4}$
• C. $\frac{\sqrt2-\sqrt6}{2}$
• D. $\frac{\sqrt4}{4}$

Answer: (A)

Find the exact value of the expression.
$\sin15\degree$

• A. $\frac{\sqrt2-\sqrt6}{4}$
• B. $\frac{\sqrt6-\sqrt2}{4}$
• C. $\frac{\sqrt2-\sqrt6}{2}$
• D. $\frac{\sqrt4}{4}$

Answer: (B)

Find the exact value of the expression.
$\cos\frac{5\pi}{12}$

• A. $\frac{\sqrt2-\sqrt6}{4}$
• B. $\frac{\sqrt6-\sqrt2}{4}$
• C. $\frac{\sqrt2-\sqrt6}{2}$
• D. $\frac{\sqrt4}{4}$

Answer: (B)

Sum and Difference of Sine & Cosine Example 4

Hey, everyone. This is a really specific type of problem that you may come across. Here we're asked to find the exact value of the expression sine of 10 degrees times cosine of 20 degrees plus the sine of 20 degrees times the cosine of 10 degrees. Now the first time that you see a problem like this, you may be thinking, okay, I don't know the sine of 10 degrees or the cosine of 20 degrees, nor do I know the sine of 20 degrees or the cosine of 10 degrees. So what do I do here?

Well, looking at this, seeing as I have the same two angles in both my terms, and I have cosines and sines, I should notice that this looks really similar to one of my sum or difference formulas. Now here, since I have the sine times the cosine plus the sine times the cosine again, I am looking at my sine formulas here. Now here, if I had a first angle \(a\) of 10 degrees and a second angle \(b\) of 20 degrees, this is really the sine of 10 degrees plus 20 degrees. Because if I were to expand this out using my sum formula right there, I would end up right back with this original expression. But what is 10 plus 20 degrees?

Well, 10 degrees plus 20 degrees is 30 degrees, and the sine of 30 degrees is a trig value that I know using my unit circle. Now the sine of 30 degrees is simply equal to one half, and that's my final answer here. So when you see something that looks a little crazy like this, think about how you can manipulate your sum and difference formulas in order to get to a value here. Thanks for watching, and I'll see you in the next one.

Find the exact value of the expression.
$\cos80\degree\cos20\degree+\sin80\degree\sin20\degree$

• A. $-\frac12$
• B. 0
• C. $\frac12$
• D. $\frac{\sqrt3}{2}$

Answer: (C)

Sum and Difference of Sine & Cosine Example 5

Hey, everyone. So far, we've been using our sum and difference identities in order to simplify and get exact values for trig functions of specific angles. But often, you'll be asked to use your sum and difference identities to work with expressions that have variables in them. So we're going to do that here. Now doing this will allow you to simplify different expressions.

So let's jump into this first one here. We have the sine of beta plus 30 degrees, and we want to expand this expression using our difference identities and then simplify. Now here, since I have the sine of theta plus 30 degrees, I'm going to use my sum formula for sine. Now expanding that out will give me the sine of theta times the cosine of 30 degrees. Then I'm adding that together with the cosine of theta times the sine of 30 degrees.

Now the cosine and sine of 30 degrees are two values that I already know from the unit circle. Now the sine of theta and the cosine of theta are not things that I can simplify because they're just of a variable, but the cosine of 30 degrees is equal to the square root of three over two, and the sine of 30 degrees is equal to one half. So here I have the sine of theta times the square root of three over two, plus the cosine of theta times one half. Now this looks kind of weird, so let's go ahead and rearrange and simplify here. This is the square root of three over two times the sine of theta because we always want to have that coefficient in the front, and then we're adding that together with one half times the cosine of theta.

Now, we can simplify this a little bit further because these both have a denominator of two. So I can rewrite this with my one half on the outside, factored out, times a root three sine theta plus the cosine of theta and we have simplified this as much as we can using our sum and difference formulas. Now let's look at another example. Here we have the cosine of Pi over four minus theta. Now here, I want to use my difference formula for cosine.

So using that to expand this out, this gives me the cosine of Pi over four times the cosine of theta plus the sine of pi over four times the sine of theta. Now again, we know these two trig values. The cosine and the sine of pi over four are actually both the square root of two over two. So this gives me the square root of two over two times the cosine of theta plus the square root of two over two times the sine of theta. Now because these both have the same exact coefficient, I can go ahead and factor this out.

Now that will leave me with the square root of two over two times the cosine of theta plus the sine of theta. And that is my expanded out using the sum and difference formulas and then simplified as much as possible. Thanks for watching, and I'll see you in the next one.

Expand the expression using the sum & difference identities and simplify.
$\sin\left(-\theta-\frac{\pi}{2}\right)$

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

Answer: (D)

Double Angle Identities

Hey, everyone. We just learned a bunch of different trig identities. And now that we know so many trig identities, we can begin to use those that we already know in order to come up with even more trig identities. Like, if we take our sum formula but for the same two angles, we end up with what's called our double angle identities. Now your first question here might be why.

Why do we need even more trig identities? The answer is that they're just going to keep allowing us to work through more and more trig problems with ease. So here, I'm going to walk you through exactly what our double angle identities are and how to use them to continue simplifying trig expressions. Let's go ahead and get started. I mentioned that these double angle identities come from using our sum formulas but for the same two angles.

So if I take the sine of some angle theta plus that same angle theta, I end up with this expression here. Now these two terms are exactly the same now, so this simplifies to two times the sine of theta times the cosine of theta, giving me my double angle identity for sine. That sine of 2θ is equal to: 2⁢sin⁢⁡⁢(θ)⁢cos⁢⁡⁢(θ) .

We can do the exact same thing for cosine and tangent, and this ends up giving me that the cosine of 2θ is equal to: cos2⁢(θ)−sin2⁢(θ) , and the tangent of 2θ is equal to: 2⁢tan⁢⁡⁢(θ) 1−tan2⁢(θ) .

Now, looking at our expression here, the cosine squared of π/12 minus the sine squared of π/12, I don't know the cosine or sine of π/12 off the top of my head, but I do recognize this as one of my double angle identities, and specifically my cosine double angle identity, the cosine squared of some angle minus the sine squared of some angle.

So, this is really just the cosine of two times that angle. So I can rewrite this as the cosine of two times my angle here, π/12. Now, two times π/12 is just π/6, so this is really the cosine of π/6, a value that I know really well from my unit circle. So I can come to an answer rather easily that this is just equal to the square root of three over two having used my double angle identity. Now, something that you may have noticed here is that we have two other forms of our cosine double angle formula, and these are just alternate forms that come from taking this first identity here and rewriting it using our Pythagorean identities.

These alternate forms are going to help us in different scenarios to simplify different trig expressions. Now that we've seen all of these double angle identities, how do we know when to use them? Well something that might be rather obvious is that whenever our argument contains two times some angle, we should use our double angle identities because that's the exact argument of all of these identities. Now, something that might be less obvious is that whenever we recognize a part of the identity, we should go ahead and use these identities. This is kind of what we did in our example here.

We recognize that this cosine squared minus sine squared was this particular part of my cosine double angle formula. Now here, this was a bit more simple, but it's not always going to be exactly like that because remember that for our other identities, we had to recognize a bunch of different forms and rearranged versions of our identities to successfully simplify expressions and verify identities. It's going to be no different with our double angle identities. So let's go ahead and work through another example here. Here we're asked to simplify the expression but not to evaluate.

The sine of 15 degrees times the cosine of 15 degrees. Now, remember that when simplifying an expression, we want to be constantly scanning for identities. And here, even though I don't know the sine or cosine of 15 degrees, I do recognize that this looks like part of my double angle formula for sine. Specifically here, I see the sine of theta times the cosine of theta. I'm just missing that two.

So, let's see what we can do here. I'm going to go ahead and rewrite my sine double angle identity here that the sine of 2θ is equal to two times the sine of theta times the cosine of theta. Now, I only want this part of this identity. So, what can I do here? Well, if I go ahead and divide both sides by two, that two will cancel on that right side, leaving me with the sine of 2θ over two is equal to the sine of theta times the cosine of theta, which is the exact expression that I'm working with here, just with theta being equal to 15 degrees.

So with this in mind, I can use that double angle identity in order to rewrite this as the sine of two times that angle, 15 degrees, over two. Now, what is two times 15 degrees? It's just 30 degrees. So this is really the sine of 30 degrees over two, and we have successfully simplified this expression, taking it from two different trig functions to just one trig function. Now we're not asked to evaluate here, but if we were, we easily could because the sine of 30 degrees is a value that I know really well from my unit circle.

Now as we continue to learn more and more identities, it can become overwhelming to figure out how to simplify a trig expression. But with each new identity that we learn, we have a greater ability to work through the problems that are asked of us. So, now that we know our double angle identities, let's continue practicing with them. Thanks for watching, and let me know if you have any questions.

Here’s what students ask on this topic:

What are the even and odd trigonometric functions, and how are they used?

Even trigonometric functions, like cosine ($\cos$) and secant ($\sec$), satisfy the property $f_{(}-x)=f_{(}x)$. This means their graphs are symmetric about the y-axis. Odd trigonometric functions, like sine ($\sin$), tangent ($\tan$), and cosecant ($\csc$), satisfy $f_{(}-x)=-f_{(}x)$, showing symmetry about the origin. These properties simplify calculations involving negative angles. For example, $\cos (-\theta )=\cos \left(\theta \right)$ and $\sin (-\theta )=-\sin \left(\theta \right)$. Recognizing these identities helps in solving equations and simplifying expressions efficiently.

What are the Pythagorean identities, and how are they derived?

The Pythagorean identities are derived from the Pythagorean Theorem applied to the unit circle. The primary identity is ${\sin }^2\theta +{\cos }^2\theta =1$. By dividing this equation by ${\cos }^2\theta$, we get ${\tan }^2\theta +1={\sec }^2\theta$. Similarly, dividing by ${\sin }^2\theta$ gives $1+{\cot }^2\theta ={\csc }^2\theta$. These identities are used to simplify expressions and solve equations involving squared trigonometric functions.

How do you use sum and difference identities to simplify trigonometric expressions?

Sum and difference identities are used when the argument of a trigonometric function involves the sum or difference of two angles. For example, the sine sum identity is $\sin (A+B)=\sin \left(A\right)\cos \left(B\right)+\cos \left(A\right)\sin \left(B\right)$. To simplify $\sin (75°)$, rewrite it as $\sin (45°+30°)$. Using the identity, this expands to $\sin (45°)\cos (30°)+\cos (45°)\sin (30°)$. Substituting known values, the result is $\frac{\sqrt{6}+\sqrt{2}}{4}$. These identities are especially useful for non-unit circle angles.

What are double angle identities, and when should they be used?

Double angle identities simplify expressions involving twice an angle. For sine, the identity is $\sin \left(2\theta \right)=2\sin \left(\theta \right)\cos \left(\theta \right)$. For cosine, it is $\cos \left(2\theta \right)={\cos }^2\theta -{\sin }^2\theta$. For tangent, it is $\tan \left(2\theta \right)=\frac{2\tan \left(\theta \right)}{1-{\tan }^2\theta }$. Use these identities when the argument involves $2\theta$ or when simplifying expressions like ${\cos }^2\theta -{\sin }^2\theta$. For example, ${\cos }^2\left(\frac{\pi }{12}\right)-{\sin }^2\left(\frac{\pi }{12}\right)$ simplifies to $\cos \left(\frac{\pi }{6}\right)$, which equals $\frac{\sqrt{3}}{2}$.

How do you verify a trigonometric identity?

To verify a trigonometric identity, start by simplifying one or both sides of the equation to show they are equal. Use known identities like Pythagorean, sum and difference, or double angle identities. For example, to verify ${\sin }^2\theta +{\cos }^2\theta =1$, recognize it as the Pythagorean identity. For more complex cases, rewrite terms using equivalent forms. For instance, to verify $\tan \left(\theta \right)=\frac{\sin }{\left(}$, rewrite $\tan \left(\theta \right)$ as $\frac{\sin }{\left(}$. Simplify step by step, ensuring each transformation is valid. Always aim to match the target expression exactly.