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.
Table of contents
- 0. Review of College Algebra4h 43m
- 1. Measuring Angles39m
- 2. Trigonometric Functions on Right Triangles2h 5m
- 3. Unit Circle1h 19m
- 4. Graphing Trigonometric Functions1h 19m
- 5. Inverse Trigonometric Functions and Basic Trigonometric Equations1h 41m
- 6. Trigonometric Identities and More Equations2h 34m
- 7. Non-Right Triangles1h 38m
- 8. Vectors2h 25m
- 9. Polar Equations2h 5m
- 10. Parametric Equations1h 6m
- 11. Graphing Complex Numbers1h 7m
6. Trigonometric Identities and More Equations
Sum and Difference Identities
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Sum and Difference Identities practice set
