Sum and Difference Identities - Online Tutor, Practice Problems & Exam Prep

Understanding sum and difference identities is crucial for simplifying trigonometric expressions and finding exact values of functions. For sine, the identity is sin(a+b) = sin(a)*cos(b) + cos(a)*sin(b). For cosine, it’s cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b). These identities facilitate solving complex problems involving angles not on the unit circle.

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 pi over 2 plus pi over 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 pi over 2 plus pi over 6, 2 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, this 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 pi over 2 plus pi over 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 pi over 2 times the cosine of that second angle pi over 6, plus the cosine of that first angle, again pi over 2 times the sine of pi over 6. Now from our knowledge of the unit circle, we can break this down even further because we know that the cosine of pi over 2 is just 0, so this entire term will go away. Then we also know that the sine of pi over 2 is equal to 1. So this will leave us with just the cosine of pi over 6, which is much easier to work with than the sine of pi over 2 plus pi over 6.

Now again, from our knowledge of the unit circle, we know that the cosine of pi over 6 is equal to the square root of 3 over 2 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 pi over 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 pi over 2 plus pi over 6. Now it 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 pi over 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 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 2 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. So the 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 substituting those values in, the cosine of 45 degrees is equal to the square root of 2 over 2, the cosine of 30 degrees is equal to the square root of 3 over 2, and then I'm adding that together with the sine of 45 degrees which is again the square root of 2 over 2 times the sine of 30 degrees, which is simply 1/2.

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 2 over 2 times the square root of 3 over 2 gives me the square root of 6 over 4. Then I'm adding that together with these two fractions multiplied by each other, which will give me the square root of 2 over 4. Now these both already have a common denominator, so I can simply rewrite this as the square root of 6 plus the square root of 2 all over 4, 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.

Example 1

Hey everyone. In this problem, we're asked to rewrite each argument as the sum or difference of 2 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 trig values of 75 degrees off the top of your head. So, 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.

Let's look at this first example. Here we have 75 degrees, and we want to rewrite that as the sum or difference of angles on our unit circle. I'm going to focus on angles in the first quadrant of the unit circle because those are the angles that I'm most familiar with. I'm going to consider 30 degrees, 45 degrees, and 60 degrees, and I want to find some combination of 2 of these angles that add or subtract to 75 degrees. Looking at these angles, the first thing I notice is that if I have 30 degrees and add that together with 45 degrees, that gives me the 75 degrees I'm looking for. Here, 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 2 out of these 3 angles that either add or subtract to that negative 15 degrees. Looking at these, I have my 45 degrees and my 30 degrees. If I subtract 45 from 30, that would give me negative 15. Here, if I took the cosine of 30 degrees minus 45 degrees, 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 \( \frac{7 \pi}{12} \). \( \frac{7 \pi}{12} \) is in radians, so this can be a little trickier to work with because it's a fraction. But remember, our angles are still these three angles: \( \frac{\pi}{6} \), \( \frac{\pi}{4} \), and \( \frac{\pi}{3} \). We want to find 2 of these angles that add or subtract to give us \( \frac{7 \pi}{12} \). Looking at \( \frac{7 \pi}{12} \), I wanna think of numbers that add or subtract to 7. First thinking of 3 and 4, here, if I took \( \frac{3 \pi}{12} \) and added it together with \( \frac{4 \pi}{12} \), that gives me \( \frac{7 \pi}{12} \). But these don't look quite like these angles yet, but let's simplify these. \( \frac{3 \pi}{12} \) can be simplified to \( \frac{\pi}{4} \) and \( \frac{4 \pi}{12} \) can be simplified to \( \frac{\pi}{3} \), which are 2 of my 3 angles here. So here, I could take the cosine of \( \frac{\pi}{4} + \frac{\pi}{3} \), then use my sum formula to get an answer.

Earlier, we saw that we want to use our sum and difference identities whenever we see angles that are multiples of 15 degrees or \( \frac{\pi}{12} \) 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. \( \frac{7 \pi}{12} \) is a multiple of \( \frac{\pi}{12} \). 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 any 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. $\frac{\sqrt4}{4}$

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. $\frac{\sqrt4}{4}$

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. $\frac{\sqrt4}{4}$

Example 2

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 of 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 1/2, 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. $\frac{\sqrt3}{2}$

Example 3

Hey, everyone. So far, we've been using our sum and difference identities 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 β + 30 degrees and we want to expand this expression using our sum and difference identities and then simplify. Now here since I have the sine of θ + 30 degrees, I'm going to use my sum formula for sine. Now expanding that out will give me the sine of θ times the cosine of 30 degrees. Then I'm adding that together with the cosine of θ 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 θ and the cosine of θ are not things that I can simplify because they're variables, but the cosine of 30 degrees is equal to the square root of 3 over 2, and the sine of 30 degrees is equal to 1/2. So here I have the sine of θ times the square root of 3 over 2, plus the cosine of θ times 1/2.

Now, this looks kind of weird, so let's go ahead and rearrange and simplify here. This is the square root of 3 over 2 times the sine of θ because we always want to have that coefficient in the front and then we're adding that together with a 1/2 times the cosine of θ. Now we can simplify this a little bit further because these both have a denominator of 2. So I can rewrite this with my 1/2 on the outside factored out times a root three sine θ plus the cosine of θ. And we have simplified this as much as we can using our sum and difference formula.

Now let's look at another example. Here we have the cosine of π/4 - θ. Now here I want to use my difference formula for cosine. So using that to expand this out, this gives me the cosine of π/4 times the cosine of θ plus the sine of π/4 times the sine of θ. Now again, we know these 2 trig values. The cosine and the sine of π/4 are actually both the square root of 2 over 2. So this gives me the square root of 2 over 2 times the cosine of θ plus the square root of 2 over 2 times the sine of θ. 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 2 over 2 times the cosine of θ plus the sine of θ. 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. $-\cos\theta$

Sum and Difference of Tangent

Hey, everyone. We just learned our sum and difference identities for sine and cosine, so we of course can't forget about tangent. Now, these identities are not going to look quite as nice, but we're still going to use them for the same purpose: to simplify expressions and to find the exact value of functions. So let's not waste any time here and jump right into our sum and difference identities for tangent. Now for the tangent of a plus b, this is going to be equal to:

tan ( a ) + tan ( b ) 1 - tan ( a ) ⋅ tan ( b )

Then for the tangent of a minus b, this is going to be almost identical except now we have:

tan ( a ) - tan ( b ) 1 + tan ( a ) ⋅ tan ( b )

Now here you'll notice that whenever we're taking the tangent of a plus b, we're going to be adding in the numerator and subtracting in the denominator. Then when we're taking the tangent of a minus b, we are instead subtracting in the numerator and adding in the denominator. Now let's go ahead and take a look at this example here and apply these new identities. So here we have the tangent of pi + pi over 4. So we want to go ahead and use our sum formula here since these angles are being added. So expanding this out, this gives me:

tan ( π ) + tan ( π 4 ) 1 - tan ( π ) ⋅ tan ( π 4 )

The tangent of pi is simply equal to 0. So that term goes away in the numerator, and then in my denominator, this entire term goes away as well because it's the tangent of pi multiplying the tangent of pi over 4. Now all I'm left with in that numerator is the tangent of pi over 4 and in my denominator just 1. But the tangent of pi over 4 over 1 is just the tangent of pi over 4. And I know from my unit circle again that the tangent of pi over 4 is simply equal to 1, giving me my final answer: The tangent of pi + pi over 4 is equal to 1, having used my sum identity there. Now remember from our sine and cosine sum and difference identities that we want to use these whenever our argument contains a plus or a minus. And we also want to use these whenever our argument contains an angle that's a multiple of 15 degrees or pi over 12 radians in order to find the exact value for trig functions that are not on the unit circle. Now remember when working with these identities, we are also going to come across expressions that have variables in them. So let's go ahead and take a look at this example here. Here we have the tangent of theta minus 45 degrees. So let's go ahead and expand this out using our difference formula since these angles are being subtracted. Here expanding this out I end up with:

tan ( θ ) - tan ( 45 ° ) 1 + tan ( θ ) ⋅ tan ( 45 ° )

The tangent of 45 degrees from our unit circle is just equal to 1. Here I can replace both of those with ones. Now all I have in my numerator is the tangent of theta minus 1. Then in my denominator, I have 1 plus the tangent of theta times 1. But the tangent of theta times 1 is just the tangent of theta, so this gives me my final simplified expression here, the tangent of theta minus 1 over 1 plus the tangent of theta. Now here you'll notice that we ended up with just a function of theta, whereas we started with a function of theta minus 45 degrees. Now, this is something that you'll be asked to do from time to time, and you can do this using your sum and difference formulas. Let's take a look at one final example here. Here we have the tangent of 90 degrees...

Find the exact value of the expression.

$\tan105\degree$

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

Expand the expression using the sum & difference identities and simplify.

$\tan\left(-\theta-\frac{\pi}{2}\right)$

• A. $-\tan\theta$
• B. $\tan\theta$
• C. $-\cot\theta$
• D. $\cot\theta$

Verifying Identities with Sum and Difference Formulas

Hey, everyone. Earlier, we used all of our fundamental trig identities in order to do what's called verifying an identity, where we were really just simplifying one or both sides of an equation with the goal of making them equal to each other. Now we know some more trig identities, our sum and difference identities, so we're going to continue to do the same thing. Nothing is changing here. We're still going to use our simplifying strategies along with our trig identities in order to make both sides of a given equation equal. So let's not waste any time here and jump right into our example.

In this example, we're asked to verify the identity using our sum and difference identities. The equation that we're given here is: cosπ2-θ=sin⁡θ. Remember that whenever we verify identities, we want to start with our more complicated side. Looking at this equation, my left side is more complicated because that argument has a difference in it.

Remembering our strategies for simplifying trig expressions, we want to be constantly scanning for identities. Working on that left side here, I see that I have a difference in that argument, which tells me that I can use my difference formula for cosine. So, using that formula, the cosine of π2-θ is cosπ2⁢cos⁢θ+sin⁢π2⁢sin⁢θ. From here, we can simplify this further because we know the cosine of π2 is just 0, so this entire term will go away. But the sine of π2 is equal to 1.

So all that I'm left with here is 1 times the sine of theta. But one times the sine of theta is just the sine of theta. Looking at the right side of my equation, remembering that my goal is for both sides of this equation to be equal, I see that the other side is already the sine of theta. So here I have that sin⁡θ=sin⁡θ and I have successfully verified this identity.

You might recognize this as being a cofunction identity that we've used earlier in this course. So here we were able to verify that cofunction identity using our sum and difference identities. Now that we know how to continue verifying identities using the new identities that we've learned, let's get some more practice. Thanks for watching, and I'll see you in the next one.

Example 4

Hey, everyone. In this problem, we're asked to verify the identity using our sum and difference identities. And the equation that we're given here is the cotangent of 90 degrees minus theta is equal to the tangent of theta. Now remember, when verifying an identity, we want to start with our more complicated side. And here, that's my left side because I see that I have a difference in that argument. So let's go ahead and get started with simplifying that left side.

Now on this left side, remember that we want to be constantly scanning for identities using our strategies over there. And the cotangent using our reciprocal identity is 1 over the tangent. So here, I can rewrite this as 1 over the tangent of 90 degrees minus theta. Now here your first instinct might be to go ahead and use your difference formula for the tangent, but let's think about what would happen if we did that. Here, since in our argument we have 90 degrees minus theta, that tells me that I'm going to end up taking the tangent of 90 degrees somewhere in that formula. But the tangent of 90 degrees is an undefined value. And remember that whenever we're using our sum and difference formulas for the tangent, if we end up with an undefined value, we actually want to go ahead and rewrite everything in terms of sine and cosine and go from there. That way, we don't end up with an undefined value.

So here, instead of rewriting this cotangent as 1 over the tangent and then using our tangent formula, we're going to instead rewrite the cotangent in terms of sine and cosine. So let's go ahead and do that here. We're going to start from scratch, not with our tangent, and we're going to take this cotangent and break it down in terms of sine and cosine so that we don't end up with an undefined value.

Now remember that the cotangent is equal to the cosine over the sine. So here I'm going to be taking the cosine of 90 degrees minus theta over the sine of 90 degrees minus theta. Now from here, I can use my sum and difference formulas for sine and cosine. So in my numerator, since I have the cosine of 90 degrees minus theta, I'm going to use my difference formula for cosine and since these are being subtracted in the argument I'm going to be adding my two terms together in that formula.

cos ( θ ) = cos ( 90° - θ ) × cos ( θ ) + sin ( 90° ) × sin ( θ )

So expanding this out gives me the cosine of 90 degrees times the cosine of theta plus the sine of 90 degrees times the sine of theta. Then in my denominator, I have the sine of 90 degrees minus theta. So I'm going to use my difference formula for sine.

sin ( θ ) = sin ( 90° - θ ) × cos ( θ ) - cos ( 90° ) × sin ( θ )

Now here, since we're subtracting in that argument, we're going to be subtracting our two terms in that identity. Now we have a bunch of values that we know from our knowledge of the unit circle. The cosine of 90 degrees is 0, that tells me that these entire terms will go away because they are both 0. And the sine of 90 degrees is simply equal to 1. So I'm just left with the sine of theta in that numerator, which is this term right here, over the cosine of theta, which is that term that's left in the denominator.

Now what is the sine of theta over the cosine of theta? Well, it's simply equal to the tangent of theta. Now looking back at that right side of my equation, remembering that my goal is for both of these sides to be equal, I see that that right side is already the tangent of theta. So pulling that all the way down here, I have successfully verified this identity, showing that the tangent of theta is indeed equal to the tangent of theta having expanded this out using my sum and difference formulas. So now we have successfully verified this identity, which you might recognize as being one of our cofunctional identities that we've used earlier in this course. Let me know if you have any questions, and thanks for watching.

Example 5

Hey everyone. In this problem, we're asked to verify the identity. And the equation that we're given here is the cosine of a plus b over the cosine of a times the cosine of b, and that's equal to the negative tangent of a times the tangent of b plus 1. Now remember that there are always going to be multiple ways to go about verifying an equation being equal, that's totally fine. Feel free to try this on your own and then check back in with me. Here, I'm gonna show you how I would do this. Now remember that when verifying an identity, we want to start with our more complicated side. And here, this fraction appears more complicated to me than this right side, so I'm going to start on that left side with that fraction. Now what do we want to do here? Well, remember one of our strategies tells us that we want to be constantly scanning for identities. And the first thing that I notice here is that in my numerator, I have the cosine of a plus b.

cos a + b cos × cos − sin × sin cos × cos

Now since we're adding in that argument, we're gonna be subtracting those two terms from each other. So let's go ahead and expand this out using that sum formula. So I'll end up with the cosine of a times the cosine of b minus the sine of a times the sine of b. Now that denominator stays the same here, the cosine of a times the cosine of b. Now we always want to be constantly checking in with what's actually going on in our expression. And looking at this expression, one thing that I notice is that in my numerator, I have this term, the cosine of a times the cosine of b. And that's the exact same term that's in my denominator. So I'm gonna do something here that seems a little bit backwards because typically we want to go ahead and combine all of our fractions. But here, since these terms are the same, I'm actually going to break this fraction up. So let's see what happens when I do that. Now breaking this fraction up, I have the cosine of a times the cosine of b over the cosine of a times the cosine of b minus the sine of a times the sine of b over the cosine of a times the cosine of b. So what happened here? Well, looking at this term, these are the exact same term in the numerator and the denominator. So this is just equal to 1. Then looking at this term over here, let's break this up further and see what happens.

1 − sin cos × sin cos

So here, I'm subtracting the sine of a over the cosine of a, and this is being multiplied by the sine of b over the cosine of b. Now what is sine of something over the cosine of that same something? Well, it's the tangent. So here, I really have a 1 minus the tangent of a times the tangent of b. So why did I want to get here? Well, remember what the right side of my equation is, because our goal here is to make both sides of our equation equal. Now that right side of my equation is negative tangent of a times the tangent of b plus 1, and that's exactly what this is if I rearrange it. So let's rearrange this and fully verify this identity. Now rearranging this to look like that right side, this is negative tangent of a times the tangent of b plus 1. Now this is exactly the same as that right side, and I don't have to do anything else to that right side. I don't have to simplify anything. I can just bring it right down here, the tangent of a times the tangent of b plus 1 and we're fully done here. We have successfully verified this identity with both sides of our equation being equal here. Let me know if you have any questions, and thanks for watching.

Example 6

Hey, everyone. In this problem, we're asked to verify the identity. And the equation that we're given here is the sine of a minus b over the tangent of a times the tangent of b is equal to the cosine of a times the cosine of b times the cotangent of b minus the cotangent of a. Now remember that there are multiple ways to verify an identity, so if you do this differently but you still end up with both sides of your equation being equal, that's totally fine. You should definitely try this problem on your own and then check back in with me. Now remember that when verifying an identity, we want to start with our more complicated side first. Now it's not always going to be super obvious which side is more complicated, but to me, fractions are always more complicated. So I'm going to start on the side that has this fraction, the sine of a minus b over the tangent of a times the tangent of b. So let's get started in simplifying this side. Now remember, we want to be constantly scanning for identities here. And here, the first thing I notice on that left side is that I have the sine of a minus b, which means that I can use my difference formula for sine. Now since we're subtracting in that argument, we're also going to be subtracting our two terms whenever we expand this out. So let's go ahead and expand this. Expanded out, this numerator is the sine of a times the cosine of b minus the cosine of a times the sine of b. Then in my denominator, I have the tangent of a times the tangent of b. Now where do we go from here? Well, looking at this, I have a bunch of sines and cosines in my numerator, but I have tangents in my denominator. Now one of our strategies tells us to break down everything in terms of sine and cosine. So let's go ahead and try that here, breaking down this denominator in terms of sine and cosine. Now remember that the tangent is equal to sine over cosine. So we're going to make this a little bit more messy before it gets more simple, but that's totally fine here. So I'm going to keep that numerator the same for now, the sine of a times the cosine of b minus the cosine of a times the sine of b. Then I am breaking down these tangents in terms of sine and cosine. So this is the sine of a over the cosine of a times the sine of b over the cosine of b. Now, where do we go from here? Because this does not look more simple. But let's think about what's happening here. Now looking at what's in my numerator, the sine of a times the cosine of b minus the cosine of a times the sine of b. And in my denominator, I have all of those same terms, but they're kind of just mixed up. So what can we do here to actually get stuff to cancel? Well, I'm going to do something that's a little bit unconventional because I'm seeing that there's a way that there are things that could cancel out. Now here, I'm going to go ahead and break up these fractions from each other. So I'm kind of doing the opposite of this one strategy, and sometimes that's going to help us out. Now let's go ahead and do that here. So breaking up this fraction, I end up with the sine of a times the cosine of b over the sine of a over the cosine of a times the sine of b over the cosine of b. Now remember that you want to stay organized here, so make sure that you can tell exactly what's going on, that this is a term over some more fractions. Now this is subtracting over the cosine of a times the sine of b, and this is divided by that same denominator, the sine of a over the cosine of a times the sine of b over the cosine of b. Now what's happening here and how do we get stuff to cancel? Well, here looking at this, I see that there are multiple things that could cancel. So let's go ahead and do that here. Now I have the sine of a on the top and the sine of a on the bottom. So those definitely cancel. Then looking at my other term, I have the sine of b on the top and the sine of b on the bottom. But I still have these other fractions. So what am I going to do here? Well, remember that we can, whenever we're dividing by a fraction, we're really just multiplying it by the reciprocal. So this can be rewritten as the cosine of b times the reciprocal of that fraction that's on the bottom. So I'm pulling these to the top, the cosine of a times the cosine of b over the sine of b. Then doing the same thing to this fraction here, the cosine of a times the cosine of a times the cosine of b, this denominator here, over the sine of a. Now, let's pause here and remember what our ultimate goal is. Remember that our goal is for both sides of our given equation to be equal. So let's look back at that other side of our equation. Now the other side of our equation is the cosine of a times the cosine of b times the cotangent of b minus the cotangent of a. So let's think about where we are and think about how we can get there because if we can get this left side to be exactly equal to this right side, then we won't have to do any simplification over on that right side. So coming back down to where we are, I see this term, the cosine of a times the cosine of b, in both terms that I'm working with here. So I can go ahead and factor that out. Now here, factoring that out, I have the cosine of a times the cosine of b, and that's multiplying the cosine of b over the sine of b minus the cosine of a over the sine of a. So what's going on here? Well, I have this exact term that is on that other side of my equation that I want this to look like. What is the cosine over the sine? Well, the cosine over the sine is the cotangent. So let's go ahead and simplify this further. I'm keeping that factor the same, the cosine of a times the cosine of b, but I am replacing each of these terms with the cotangent. Now here, I have the cotangent of b minus the cotangent of a. So let's look back at what that right side of our equation was. It is the cosine of a times the cosine of b times the cotangent of b minus the cotangent of a, which is exactly what we've ended up with here. So if I pull that right side of my equation all the way down here, I end up with the same exact term without having to do any simplification over there. So you have to get creative whenever you're verifying identities because sometimes you might have to do some things that seem unintuitive because there's a bunch of different things going on, and you always want to remember what your end goal is here. So here, we have successfully verified this identity, and we are done. Thanks for watching, and let me know if you have any questions.

Evaluating Sums and Differences Given Conditions

Hey, everyone. As you continue to work through problems dealing with your sum and difference identities, you're going to come across a very specific type of problem in which you'll be given trig values and asked to evaluate a sum or a difference based on those. But you won't actually be given the value of any angles. Now this sounds kind of strange and, like, it might be kind of complicated. But you actually already have all of the knowledge and information that you need to solve these types of problems. But it can still be easy to get lost in all of this information. So here, I'm going to break down for you step by step exactly how to solve these sorts of problems and how exactly to interpret all of this information. So let's go ahead and get started. Now looking at our example problem here, we are told to find the sine of a plus b. Given that the cosine of a is equal to 4/5, the sine of b is equal to 5/13, and angle a is in quadrant 4, and angle b is in quadrant 2. Now what do we do with all of that information? Well, let's take a look at our steps here.

Now, in step 1, we are told to expand our identity and then identify any unknown trig values. So let's take a look at what we're doing here. Here, we're trying to find the sine of a+b. Now, this is my sum identity for sine, so I'm going to go ahead and expand that identity out. Now doing that, I end up with sina+b=sina∙cosb+cosa∙sinb. Now, I've done the first part of that step 1, and I've expanded that identity. But now we want to go ahead and identify any of our unknown trig values. Now in my problem, I'm given that the cosine of a is equal to 4/5 and the sine of b is equal to 5/13. So I already know the cosine of a and the sine of b in that expanded identity. So my unknown trig values here are going to be the sine of a and the cosine of b. Those are my unknown trig values. Now that I've completed step number 1, we can move on to step number 2. Now our goal here is to eventually find these missing trig values, so let's continue on in our steps to get there.

Now, in step 2, we are told that from our given info, we want to sketch and label right triangles in the proper quadrant. Now it's important here to pay close attention to the sign of all of our values because they may be different depending on what quadrant our angle is located in. Now here, we're told that angle a is in quadrant 4. So I'm going to go ahead and draw my right triangle and label that angle in quadrant 4, angle a. Now, I'm also told that the cosine of a is equal to 4/5. So using that information, that tells me that my adjacent side is equal to 4 and my hypotenuse is equal to 5. Now both of these values are going to be positive because our hypotenuse will always be positive and this side length here is in the positive x values. Now let's go ahead and label our other triangle. We're told the angle b is in quadrant 2, so I'm going to go ahead and draw another right triangle here in quadrant 2 and label my angle b. Now we're told that the sine of b is equal to 5/13, which tells me that my opposite side is equal to 5 and my hypotenuse is equal to 13. So here we have successfully completed step number 2 having sketched and labeled both of our triangles here.

Now moving on to step number 3, we want to find any missing sides using our Pythagorean theorem. Now looking at my first triangle here, I am missing this one side length. And looking at these values, I actually don't have to use the Pythagorean theorem because I recognize that this is a 3, 4, 5 triangle. So this missing side length is simply 3. But remember, we want to continue paying attention to the sign of our values here. And since this is in the negative y values, this is actually a negative 3. Now let's take a look at our other triangle. We're missing this side length here. Now you may go ahead and use the Pythagorean theorem here, but here I also notice that this is a special right triangle, so this is a 5, 12, 13 triangle. So this missing side length is simply 12. But again, this is a negative 12 because here we're in the negative x values. Now we have completed step number 3. We have found those missing side lengths. And from here, we can solve for unknown trig values that we identified in step number 1.

Now in step 1, we said that our unknown trig values were the sine of a and the cosine of b. So let's go ahead and solve for those. Now for our first triangle, we want to find the sine of this angle a. Now the sine of angle a is going to be that opposite side of negative 3 divided by my hypotenuse of 5. Now in my other triangle here, we want to find the cosine of this angle b. That's my other unknown trig value, the cosine of b. Now the cosine of b is going to be that adjacent side, negative 12, over that hypotenuse 13. So those are my 2 missing trig values. Now from here, we can finally move on to our last step where we can actually plug in all of those values and simplify.

Now here, we want to go ahead and plug in those values that we just found in step number 4, negative threefifths and negative twelve thirteenths. So here, plugging that in, negative 3/5 for the sine of a multiplying that by the cosine of b, negative 12/13. Now I'm adding that together with those trig values that I was already given in my problem statement. The cosine of a, 4/5 that we see up in our problem here, and then the sine of b, which is 5/13 also in that problem statement, 5/13. Now from here it's just algebra and we just need to simplify. Now because I have these fractions, I need to go ahead and multiply them across in order to simplify this further. Now looking at that first fraction, negative 3/5 times negative 12/13, multiplying across gives me a positive 36/65. Then I'm adding that together with my other fraction multiplied across, which is going to give me 20/65. Now I can go ahead and add these two fractions together because they already have a common denominator. So adding these two fractions, I end up with 56/65. And this is my final answer here. The sine of a plus b given all of this information is 56/65. Now I know that these sorts of problems can be really tedious, so make sure that you're taking your time and paying attention to what you're doing in each step. Thanks for watching, and let me know if you have any questions.

Example 7

Hey, everyone. In this problem, we're asked to find the sine of a+b given that the sine of a is equal to three fifths, the sine of b is equal to negative one-half, and we're also given that angle a is between π2 and π, and angle b is between 3π2 and 2π. So let's go ahead and get started here with our steps. Now here we're told to expand our identity and then identify any unknown trig values. Now here we're trying to find the sine of a+b. So we of course want to use our sum identity for sine. So let's go ahead and expand that out. Now the sine of a+b is going to be equal to the sine of a times the cosine of b plus the cosine of a times the sine of b. Now which of these trig values do we already know? Well, in my problem, I'm told that the sine of a is equal to three fifths and the sine of b is equal to negative one-half. So those are my two trig values that I already know. So my missing trig values are the cosine of b and the cosine of a. Now let's get started in finding those missing trig values by continuing on in our steps. Step one is done. Moving on to step number two. From all of the info that we're given in that problem statement, we want to go ahead and sketch and label our triangles. Remember that we have to pay close attention to our signs here. Let's go ahead and start with angle a. Now we're told that angle a is between π2 and π, which is right here in quadrant 2. So I want to go ahead and sketch my right triangle and label that angle a there. Now here I have my angle a. And I also know that the sine of angle a is equal to three fifths. So that tells me that my opposite side is equal to 3 and my hypotenuse is equal to 5. Now for my angle b, I'm told that angle b is between 3π2 and 2π, which is down here in quadrant 4. So we want to go ahead and sketch our triangle in quadrant 4 and then label that angle b. Now here I have my angle b and I also know that the sine of this angle is equal to negative one-half. Now that tells me that my opposite side is equal to negative one. This is a negative value because it's in the negative quadrant here for my y values. Then my hypotenuse is equal to 2, which remember is always going to be positive. So seeing as the sine value is negative one-half, that opposite side is negative one, that hypotenuse is 2. Now we have completed step number two. Moving on to step number three, we want to find any of our missing side lengths. So let's start with our triangle right here with angle b. Now here we're missing this side. So I'm going to set up my Pythagorean Theorem, a2 plus b2 equals c2. Now using my Pythagorean Theorem, I am missing one of my side lengths. So I'm going to keep a as that variable that I'm solving for. Then I'm adding that together with b2, which is negative one, and that's equal to c2, which is 2. So squaring those values gives me a2 plus 1 is equal to 4. Now if I subtract 1 on both sides, it cancels out, leaving me with a2 is equal to 3. So my final answer here is that a is equal to the square root of 3. Now you might have already noticed that that side length would be the square root of 3 because of those other two values. So if you already recognize that, that's totally fine. You don't have to go through using the Pythagorean Theorem unless you need to. Now let's look at our other triangle here. Here for my angle a, I'm missing this side length here, but my other two side lengths are 3 and 5. So this is a 3, 4, 5 triangle, and that missing side length is 4. But remember, we got to pay attention to that sign. And this is a negative 4 because of where it's located on my coordinate system. Now we've completed step number three. We can actually solve for those unknown trig values. Now our unknown trig values here, remember, were the cosine of b and the cosine of a. So let's go ahead and solve for those starting with the cosine of b. Now the cosine of b using this triangle here is going to be equal to that adjacent side, the square root of 3, over that hypotenuse of 2. Then for the cosine of a using my other triangle, the cosine of a is going to be equal to again that adjacent side negative 4 over that hypotenuse 5. Now we have all of the information that we need to actually evaluate the sum. So let's move on to step number five and finally plug in all of our values. Now doing that here, remember that the sine of a and the sine of b were already in our problem statement. We already know what they are. So the sine of a was given as three fifths. And then the cosine of b is what I just found. It is root 3 over 2. And then I'm adding that together with the cosine of a, another value that I just found, negative four-fifths, and then multiplying that by the sine of b, which was again in my problem statement as negative one-half. So now we have all of our values here, and all that's left to do is algebra. So let's go ahead and start multiplying. Now we wanna multiply these two fractions here and doing that gives me 3√3/10. Then I'm adding that together with these two fractions multiplied across. Negative four-fifths times negative one-half gives me positive 4/10. Now again, because these have a common denominator, I can go ahead and just add those across. So this is going to give me 33+410, and this is my final answer here. Please let me know if you have any questions. Thanks for watching.

Find $\cos\left(a+b\right)$ given $\cos a=\frac12$, $\sin b=\frac12$, & $a$ is in Q IV and $b$ is in Q II.

• A. 0
• B. $\frac{\sqrt3}{4}$
• C. 1
• D. $-\frac{\sqrt3}{2}$

Here’s what students ask on this topic:

What are the sum and difference identities for sine and cosine?

The sum and difference identities for sine and cosine are essential for simplifying trigonometric expressions and finding exact values of functions. For sine, the identities are:

$\sin (a+b)=\sin \left(a\right)\bullet \cos \left(b\right)+\cos \left(a\right)\bullet \sin \left(b\right)$

$\sin (a-b)=\sin \left(a\right)\bullet \cos \left(b\right)-\cos \left(a\right)\bullet \sin \left(b\right)$

For cosine, the identities are:

$\cos (a+b)=\cos \left(a\right)\bullet \cos \left(b\right)-\sin \left(a\right)\bullet \sin \left(b\right)$

$\cos (a-b)=\cos \left(a\right)\bullet \cos \left(b\right)+\sin \left(a\right)\bullet \sin \left(b\right)$

How do you use sum and difference identities to find the exact value of trigonometric functions?

To find the exact value of trigonometric functions using sum and difference identities, follow these steps:

1. Identify the angles involved and express the given angle as a sum or difference of known angles.

2. Use the appropriate sum or difference identity. For example, for sine:

$\sin (a+b)=\sin \left(a\right)\bullet \cos \left(b\right)+\cos \left(a\right)\bullet \sin \left(b\right)$

3. Substitute the known values from the unit circle or other sources.

4. Simplify the expression to find the exact value.

For example, to find $\cos (15°)$, express it as $\cos (45°-30°)$ and use the identity:

$\cos (a-b)=\cos \left(a\right)\bullet \cos \left(b\right)+\sin \left(a\right)\bullet \sin \left(b\right)$

Substitute the values and simplify to get the exact value.

When should you use sum and difference identities in trigonometry?

Sum and difference identities should be used in trigonometry when:

1. The argument of a trigonometric function contains a sum or difference of angles, such as $\sin (a+b)$ or $\cos (a-b)$.

2. You need to find the exact value of a trigonometric function for an angle that is not on the unit circle. For example, finding $\cos (15°)$ by expressing it as $\cos (45°-30°)$.

3. Simplifying trigonometric expressions that involve multiple angles to make them easier to work with.

4. Verifying trigonometric identities by transforming one side of an equation to match the other side using these identities.

These identities are particularly useful for angles that are multiples of 15 degrees or $\pi /12$ radians.

How do you verify trigonometric identities using sum and difference identities?

To verify trigonometric identities using sum and difference identities, follow these steps:

1. Identify the more complicated side of the equation to simplify first.

2. Look for opportunities to apply sum and difference identities. For example, if you see $\cos (a-b)$, use the identity:

$\cos (a-b)=\cos \left(a\right)\bullet \cos \left(b\right)+\sin \left(a\right)\bullet \sin \left(b\right)$

3. Substitute known values or simpler expressions to transform the complicated side.

4. Simplify the expression step-by-step, ensuring each transformation is valid.

5. Continue until both sides of the equation are identical.

For example, to verify $\cos (\mathrm{\pi }/2-\theta )=\sin \left(\theta \right)$, use the difference identity for cosine and simplify using known values from the unit circle.

What are the sum and difference identities for tangent?

The sum and difference identities for tangent are used to simplify expressions and find exact values of functions. They are:

For the tangent of a sum:

$\tan (a+b)=\frac{\tan \left(a\right)+\tan \left(b\right)}{1-\tan \left(a\right)\bullet \tan \left(b\right)}$

For the tangent of a difference:

$\tan (a-b)=\frac{\tan \left(a\right)-\tan \left(b\right)}{1+\tan \left(a\right)\bullet \tan \left(b\right)}$

These identities are particularly useful when dealing with angles that are not on the unit circle or when simplifying complex trigonometric expressions.