Sum and Difference Identities - Video Tutorials & Practice Problems

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1

concept

Sum and Difference of Sine & Cosine

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Hey, everyone, we just learned a bunch of different trick identities that we've used to simplify trick expressions. But we're not quite done yet looking at this expression here, the sign of pi over two plus pi over six, 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 different identities. Now, here, I'm gonna walk you through exactly what our sum and different identities are and how we can use them to continue simplifying trick expressions and also to find the exact value of trig function 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 are some indifference identities are going to be useful whenever we have multiple angles in the argument of a trick function. Like here we have the sign of pi over two plus pi over 62 separate angles inside that single argument. Now we can use our sum indifference identities in order to expand this out and make it easier to work with. So looking at our first identity here, the sign of some angle A plus A 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 sign of pi over two pi over six, which you might also see written in degrees as the sign 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 sign of that first angle pi over two, the cosine of that second angle pi over six plus the cosine of that first angle again, pi over two times the sine of pi over six. Now, from our knowledge of the unit circle, we can break this down even further because we know that the cosine of pi over two is just zero. So this entire term will go away. Then we also know that the sine of pi over two is equal to one. So this will go away and leave me with just the cosine of pi over six, which is much easier to deal with. Then the sine of pi over two plus pi over six. Now again, from our knowledge of the unit circle, we know that the cosine of pi over six is equal to the square root of 3/2. 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 sign 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 six, which is much easier to work with. Now, let's take a look at our other sum and different 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 backwards to my, 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 in different 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 in different identities. Like here we saw the sign of pi over two plus pi over six. 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 pi over 12 radiant. Now, how do we recognize that? Because that seems like kind of an obscure piece of criteria that tells us to use our sum indifference 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 direct 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 and 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 function 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 in different 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 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 s of 45 degrees that first angle again times the s of 30 degrees. Now these things that I already know from the unit circle, I know all of these trick values. So replacing all of these substituting those values in the cosine of 45 degrees is equal to the square root of 2/2, the cosine of 30 degrees is equal to the square root of 3/2. And then I'm adding that together with the sine of 45 degrees, which is again the square root of 2/2 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, multi applying these fractions together the square root of 2/2 times the square root of 3/2, gives me the square root of 6/4. Then I'm adding that together with these two fractions multiplied by each other, which will give me the square root of 2/4. 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 indifference identities, let's continue practicing with them. Thanks for watching and I'll see you in the next one.

2

example

Sum and Difference of Sine & Cosine Example 1

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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 sign of 75 degrees. Now, you probably don't know the trig values of the 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 different formula in order to get to an answer much quicker. So let's look at the first example, here we have 75 degrees and we want to rewrite that as the sum or difference of angles on our unit circle. Now 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 the most familiar with. So here 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. Now, looking at these angles. The first thing that I noticed is that if I have 30 degree degrees and I add that together with 45 degrees that gives me the 75 degrees that I'm looking for. So here, if I took the sign of 30 degrees plus 45 degrees and then used a some formula I could get to my answer. Now, let's look at a second example. Here we have the cosine of negative 15 degrees. So again, we wanna find two out of these three angles that either add or subtract to that negative 15 degrees. Now 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 and 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. Now seven pi over 12 is in radiance. So this can be a little bit 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. So we want to find two of these angles that add or subtract to give us seven pi over 12. Now looking at seven pi over 12, I want to be thinking of numbers that either add or subtract to that, right? So if I think of that seven and I think of two numbers that add or subtract seven, I first think of three and four. Now here, if I took three pi over 12 and I added 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 su formula to get an answer. Now earlier, we saw that we want to use our some in different identities whenever we see angles that are multiples of 15 degrees or pi over 12 radiance. And that's exactly what we see here. Here, we have the sign of 75 degrees. 75 is a multiple of 15 as is negative 15, 7 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 different, and then use our identities from there. Thanks for watching and let me know if you have questions

3

Problem

Problem

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

$\frac{\sqrt4}{4}$

4

Problem

Problem

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

$\frac{\sqrt4}{4}$

5

Problem

Problem

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

$\frac{\sqrt4}{4}$

6

example

Sum and Difference of Sine & Cosine Example 2

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1m

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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, OK, I don't know the sign of 10 degrees or the cosine of 20 degrees nor do I know the sign 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 sign, I should notice that this looks really similar to one of my summer difference formulas. Now here, since I have the S times the cosine plus the S times the cosine. Again, I am looking at my S 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 sign of 10 degrees plus 20 degrees. Because if I were to expand this out, you my some 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 sign of 30 degrees is a trig value that I know using my unit circle. Now, the sign 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 in different formulas in order to get to a value here. Thanks for watching and I'll see you in the next one.

Hey, everyone so far, we've been using our sum and different 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 different identities to work with expressions that have variables in them. So we're gonna 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 sign of beta plus 30 degrees and we want to expand this expression using our sum in different identities and then simplify. Now here, since I have the sign of theta plus 30 degrees, I'm gonna use my sum formula for s 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 the and the cosine of the 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 3/2 and the sine of 30 degrees is equal to one half. So here I have the sine of theta times the square root of 3/2 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 3/2 times the sine of the because we always want to have that coefficient in the front. And then we're adding that together with a one half times the cosine of data. Now we 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 the. And we have simplified this as much as we can using our sum in different formula. 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 data plus the sine of pi over four times the sine of data. Now again, we know these two trig values, the cosine and the sine of pi over four are actually both the square root of 2/2. So this gives me the square root of 2/2 times the cosine of data plus the square root of 2/2 times the sine of data. Now, because these both have the same Act coefficient, I can go ahead and factor this out now, that will leave me with the square root of 2/2 times the cosine of the plus the sine of data. And that is my expanded out using the sum in different formulas and then simplify it as much as possible. Thanks for watching and I'll see you in the next one.

9

Problem

Problem

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

$-\cos\theta$

10

concept

Sum and Difference of Tangent

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4m

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Everyone, we just learned our sum indifference 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 different identities for tangent. Now, for the tangent of A plus B, this is going to be equal to the tangent of A plus the tangent of B over one minus the tangent of A times the tangent of B. Then for the tangent of A minus B, this is going to be almost identical, except now we have the tangent of A minus the tangent of B over one plus the tangent of A times the tangent of 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 that numerator and adding in that 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 plus pi over four. So we want to go ahead and use our some formula here since these angles are being added. So expanding this out, this gives me the tangent of that first angle pi plus the tangent of that second angle pi over four. And this gets divided by one minus the tangent of that first angle. Again, pi times the tangent of Pi over four. Now we know these angles from the unit circle right. And we know that the tangent of pi is simply equal to zero. 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 four. Now all I'm left with in that numerator is the tangent of Pi over four. And in my denominator just one, but the tangent of pi over 4/1 is just the tangent of pi over four. And I know from my unit circle again that the tangent of pi over four is simply equal to one. Giving me my final answer. That's gadget of pi plus pi over four is equal to one, having used my sum identity there. Now remember from our sine and cosine sum sum in different 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 radiance 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. So here expanding this out, I end up with the tangent of that first angle theta minus the tangent of that second angle 45 degrees. And this is getting divided by one plus the tangent of theta times the tangent of 45 degrees. Now we know the tangent of 45 degrees from our unit circle even if we don't know the tangent of data because that's just a variable but the tangent of 45 degrees is just equal to one. So here I can replace both of those with once. Now all I have in my numerator is the tangent of the minus one. Then in my denominator, I have the, I have one plus the tangent of theta times one but the tangent of the times one is just the tangent of data. So this gives me my final simplified expression here, the tangent of theta minus 1/1 plus the tangent of the. 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 different formulas. Let's take a look at one final example. Here, here we have the tangent of 90 degrees plus theta. So let's go ahead and use our sum formula to expand this out. Here. I end up with a tangent of 90 degrees plus the tangent of the over one minus the tangent of 90 degrees times the tangent of data. Now, at this point, you might be thinking, OK, I know the tangent of 90 degrees from my unit circle. But thinking of what that value actually is, the tangent of 90 degrees is undefined. So here we would end up with undefined plus the tangent of the over one minus another undefined value. So how do we do with that here? Well, whenever you end up with a tangent of A or the tangent of B being undefined, you're going to start over and you're going to rewrite the tangent of A plus or minus B as the sign over the cosine. So here with the tangent of 90 degrees plus the, I would rewrite this as the sign of 90 degrees plus the over the cosine of 90 degrees plus beta. And then go from there using my sum indifference formulas for sine and cosine rather than for tangent. Because that way I won't end up with an undefined value. Now that we know how to use our sum indifference identities for tangent. Let's continue practicing. Thanks for watching and let me know if you have questions.

11

Problem

Problem

Find the exact value of the expression.

$\tan105\degree$

A

0

B

$\sqrt3-2$

C

$-2-\sqrt3$

D

$\frac{\sqrt3+2}{2}$

12

Problem

Problem

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

$\cot\theta$

13

concept

Verifying Identities with Sum and Difference Formulas

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2m

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Hey, everyone earlier, we used all of our fundamental trig identities in order to do what's called verify 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 trick identities are some indifference identities. So we're going to continue to do the same thing. Now, 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. Now, in this example, we're asked to verify the identity using our sum in different identities. And the equation that we're given here is the cosine of pi over two minus the is equal to the sign of the. Now remember that whenever we verify identities, we want to start with our more complicated side and looking at this equation, my left side is more complicated because that argument has a difference in it. Now remembering our strategies for simplifying trig expressions, we want to be constantly scanning for identities. So 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, since we are subtracting in that argument, we're going to be adding our two terms together in that identity. So let's go ahead and expand this out using that difference identity. So the cosine of pi over two minus theta is the cosine of pi over two times the cosine of theta plus the sine of pi over two times the sine of the. Now, from here, we can simplify this further because we know the cosine of pi over two and the sine of pi over two from our knowledge of the unit circle. Now, the cosine of pi over two is just zero. So this entire term will go away but the sine of pi over two is equal to one. So all that I'm left with here is one times the sine of theta, but one times the sine of theta is just the s of theta. Now, 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 that other side is already the sign of data. So here I have that the sign of data is equal to the sign of data and I have successfully verified this identity. Now, you might recognize this as being a coun identity that we've used earlier in this course. So here we were able to verify that coun identity using our sum indifference 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.

14

example

Verifying Identities with Sum and Difference Formulas Example 4

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Hey, everyone in this problem, we're asked to verify the identity using our sum in different identities. And the equation that we're given here is the cotangent of 90 degrees and minus the is equal to the tangent of data. 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 standing for identities using our strategies over there. And the cotangent using our reciprocal identity is one over the tangent. So here I can rewrite this as one over the tangent of 90 degrees minus theta. Now, here your first instinct might be to go ahead and use your difference formula for 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 t 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 indifference formulas for tangent, if we end up with an undefined value, we actually want to go ahead and rewrite everything in terms of sign and cosign and go from there. That way we don't end up with an undefined value. So here, instead of rewriting this cotangent as 1/10 tangent, and then using our tangent formula, we're going to instead rewrite the code tangent 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 the over the sine of 90 degrees minus the. Now, from here, I can use my sum in different formulas for sine and cosine. So in my numerator, since I have the cosine of 90 degrees and minus the, 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. So expanding this out gives me the cosine of 90 degrees times the cosine of theta plus the sign of 90 degrees times the sign of the. Then in my denominator, I have the sign of 90 degrees minus the. So I'm going to use my difference formula for sign. Now here, since we're subtracting in that argument, we're going to be subtracting our two terms in that identity. So expanding that sign out using my difference formula there, I'm going to get the sign of 90 degrees times the cosine of theta minus the cosine of 90 degrees times the sign of data. So let's see what happens here. We have a bunch of values that we know from our knowledge of the unit circle. The cosine of 90 degrees is and the sign of 90 degrees. Now the cosine of 90 degrees is zero. So that tells me that these entire terms will go away because they are both zero and a sign of 90 degrees is simply equal one. So I'm just left with the sign of data in that numerator, which is this term right here over the cosine of data, which is that term that's left in the denominator. Now, what is the sign of data over the cosine of data? Well, it's simply equal to the tangent of data. Now, looking back to 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 data. So pulling that all the way down here, I have successfully verified this identity showing that the tangent of data is indeed equal to the tangent of data having expanded this out using my sum indifference formulas. So now we have successfully verified this identity, which you might recognize as being one of our coun identities that we've used earlier. In this course, let me know if you have any questions and thanks for watching.

15

example

Verifying Identities with Sum and Difference Formulas Example 5

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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 one. Now remember that there are always going to be multiple ways to go about verifying 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, feel free to try this on your own and then check back in with me here. I'm going to 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? Or 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, which tells me that I can use my sum formula for the cosine. Now, since we're adding in that argument, we're going to be subtracted those two terms from each other. So let's go ahead and expand this out using that some 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 noticed is that in my numerator, I have this term, the cosine of times the cosine of B and that's the exact same term that's in my denominator. So I'm going to 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 while looking at this term? These are the exact same term in the numerator and the denominator. So this is just equal to one. Then looking at this term over here, let's break this up further and see what happens. So here, I'm subtracting the sign 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 cosine of that? Say some same something? Well, it's the tangent. So here I really have one 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 a negative tangent of A times tangent of B plus one. 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 one. 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 one. 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.

16

example

Verifying Identities with Sum and Difference Formulas Example 6

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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 is over the tangent of A times the tangent of B is equal to the cosine of A times the cosine of B times the code tangent 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 sign of A minus B over the tangent of A times A tangent of B. So let's get started and simplify this side. Now, remember we want to be constantly scanning for identities here and here. The first thing I noticed on that left side is that I have the sign of A minus B, which means that I can use my difference formula for sign. 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 gonna 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 that other term, 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 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 sign of A on the top and the sign of A on the bottom. So those definitely cancel. Then looking at my other term, I have the sign of B on the top and the sign of B on the bottom, but I still have these other fractions. So what am I gonna do here? Well, remember that we can do 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 sign 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 S, 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 questions.

17

concept

Evaluating Sums and Differences Given Conditions

Video duration:

6m

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Hey, everyone, as you continue to work through problems dealing with your sum in different 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 in order 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 sign of A plus B given that the cosine of A is equal to 4/5 the sign of B is equal to 5/13 and angle A is in quadrant four and angle B is in quadrant two. Now, what do we do with all of that information. Well, let's take a look at our steps here. Now, in step one, 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 sign of A plus B. Now this is my sum identity for sign. So I'm gonna go ahead and expand that identity out. Now doing that, I end up with the sign of A plus B is equal to the S of A times the cosine of B plus the cosine of A times the sine of B. Now, I've done the first part of that step one 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 S 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 identified those we have completed step number one and we can move on to step number two. Now, our goal here is to eventually find these missing trick values. So let's continue on in our steps to get there. Now, in step two, 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 four. So I'm going to go ahead and draw my right triangle and label that angle in quadrant four, angle A. Now I'm also told that the cosine of A is equal to four fits. So using that information that tells me that my adjacent side is equal to four and my hypotenuse is equal to five. Now, both of these values are going to be positive because our hypotenuse will always be positive. And this side link here is in the positive X values. Now, let's go ahead and label our other angle. We're told the angle B is in quadrant two. So I'm going to go ahead and draw another right triangle here in quadrant two and A label my angle B. Now we're told that the sign of B is equal to 5/13, which tells me that my opposite side is equal to five and my hypotenuse is equal to 13. So here we have successfully completed step number two, having sketched and labeled both of our triangles here. Now, moving on to step number three, we want to find any missing sides using our Pythagorean theorem. Now, looking at my first triangle here, I am missing this one side link and looking at these values, I actually don't have to use the Pythagorean theorem because I recognize that this is a 345 triangle. So this missing side link is simply three. 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 three. Now let's take a look at our other triangle. We're missing this side length here. Now, so you may go ahead and use the Pythagorean theorem here. But here I also noticed that this is a special right triangle. So this is a 5 1213 triangle. So this missing side link 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 three, we have found those missing side links. And from here, we can solve for our unknown trig values that we identified in. Step number one. Now, in step one, we said that are unknown trig values were the sine of A and the cosine of B. So let's go ahead and solve for this. 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 three divided by my hypotenuse of five. 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 co and B is going to be that adjacent side negative 12 over that hypotenuse 13. So those are my two 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 four, negative 3/5 and negative 12/13. So here plugging that in negative 3/5 for the sign 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 sign 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 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 common denominator. So adding these two fractions, I end up with 56/65. And this is my final answer here. The sign 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.

18

example

Evaluating Sums and Differences Given Conditions Example 7

Video duration:

5m

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Everyone in this problem, we're asked to find the sign of A plus B given that the sign of A is equal to 3/5 the S of B is equal to negative one half. And we're also given that angle A is in between pi over two and pi and angle B is in between three pi over two and two pi. 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 trick values. Now, here we're trying to find the sine of A plus B. So we of course, want to use our sum identity for sine. So let's go ahead and expand that out. Now, the S of A plus 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 3/5 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 in between pi over two and pi which is right here in quadrant two. 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 sign of angle A is equal to 3/5. So that tells me that my opposite side is equal to three and my hypotenuse is equal to five. Now for my angle B, I'm told that angle B is in between three pi over two and two pi which is down here in quadrant four. So we want to go ahead and sketch our triangle in quadrant four and then label that angle B. Now, here I have my ankle B and I also know that the sign of this angle is equal to negative one half. Now that tells me that my opposite side is equal to negative one. Now this is a negative value because it's in the negative quadrant here for my Y values, then my hypotenuse is equal to two which remember is always going to be positive. So seeing as the sign value is negative one half, the opposite side is negative one that hypotenuse is two. Now we have completed step number two, moving on to step number three, we want to find any of our missing side links. 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 A squared plus B squared equals C squared. Now using my Pythagorean theorem, I am missing one of my side links. So I'm going to keep A as that variable that I'm solving for, then I'm adding that together with B squared, which is negative one and that's equal to C squared, which is two. So squaring those values gives me a squared plus one is equal to four. Now, if I subtract one on both sides, it cancels out leaving me with A squared is equal to three. So my final answer here is that A is equal to the square root of three. Now, you might have already noticed that that side length would be the square root of three 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 from my angle A, I'm missing this side length here. But my other two side links are three and five. So this is a 345 triangle and that missing side link is four. But remember we got to pay attention to that sign and this is a negative four 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 we, 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 three over that hypotenuse of two. Then for the cosine of A using my other triangle, the sign of A is going to be equal to again that adjacent side negative four over the hypotenuse five. Now we have all of the information that we need to actually evaluate this 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 sign of A and the sign of B were already in our problem statement, we already know what they are. So the sign of A was given as 3/5 and then the cosine of B is what I just found it is route 3/2. And then I'm adding that together with the cosine of a, another value that I just found negative 4/5. 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 and we want to multiply these two fractions here and doing that gives me three root 3/10. Then I'm adding that together with these two fractions multiplied across negative 4/5 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 numerators across. So this is going to give me three root three plus 4/10 and this is my final answer here. Please let me know if you have any questions. Thanks for watching.

19

Problem

Problem

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