Learn about Trig identies part 3 - part 5 if you watch the proofs Video

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Learn about Trig identies part 3 - part 5 if you watch the proofs

Learn about Trig identities part 3 - part 5 if you watch the proofs Welcome back. I have to— because I was hitting against the You Tube 10 minute limit. But I was always saying as we said cosine of minus a so I drew a right triangle with a and then I showed minus a and I said, “Well, all of the lengths are going to be the same but now the direction of—.” We’re kind of assuming this is all unit circles. If you don’t now remember unit circle, maybe you want to re-watch the videos we have on that. But I’m just showing you that the cosine of minus a is equal to this side over the hypotenuse and this hypotenuse is the same as this hypotenuse. So cosine of minus a is adjacent over this hypotenuse but it’s the same thing so we know that cosine of minus a is equal to cosine of a. And actually by definition that makes it—I don’t want to confuse you too much but that makes cosine even functional. And I’ll show you more—actually I should do a whole presentation on even not functions. And let’s see what sine of minus a is. Sine of minus a is equal to—so this is minus a, so it’s this side—so it’s the minus length of—let’s call this x, let’s call this y and let’s call this h. And if that is x, this is y, this length is y but this length right here is minus x. So the sine of minus a is minus x over h. What’s the sine of a? Sine of a is equal to—this is a opposite of hypotenuse x over h. So sine of minus a is equal to minus 1 times x over h or this is just the same thing at—I don’t know if we could multiply both sides of this by minus 1 minus x over h. So sine of minus a is equal minus sine of a. So let me clear this out and rewrite this identity. And as you can see, all I’m doing is I’m just playing around with triangles and showing you that just using the basic Socatoa you can actually discover a whole set of Trigonometric identities. So let’s clear that and you might—it’s useful to memorize and normally advocate memorizing but it’s helpful just to do things quickly. But I’d also advocate being able to prove it to yourself so if you ever forget it and you don’t have a cheat-cheat available you can prove it and if you ever have to teach it then you’ll be I think able to explain the underlined themes a little bit better. So let’s clear this, let’s see if we can discover some more trig identities. So we know that—so let’s see if we have sine, what’s the sine of a plus pi over 2? Well, we could use our handy sine of a plus b identity which we have already proved so we can use it now. So that’s the sine of a times the cosine of pi over 2 plus the sine of pi over 2 and we’re in radiance of course I could—this could have been 90 degrees instead if we wanted to be in degrees. Sine if pi over 2 times the cosine of a. Well, this equals to sine of a—what’s the cosine of pi over 2 or cosine of 90 degrees? So that’s one. We’re on the unit circle where we’re pointing straight up so the x coordinate is zero, I could draw that out but I think you might want to draw the unit circle and figure out it for yourself or if you don’t—use a calculator but you will learn that it is zero, cosine of 5 over 2 is zero plus sine of pi over 2 for the same reason we’re pointing straight up on the unit circle so the y coordinate or the sine coordinate is 1 right on the unit—where it essentially is in the point zero 1 on the unit circle. So sine of pi over 2 is 1 so then times cosine of a. So sine of a times zero is zero, 1 times cosine of a is just cosine of a so we have a new useful trig identity. Sine of a plus pi over 2 is equal to cosine of a, fascinating. So really, this is just telling us that cosine of a is the same thing as sine of a shifted. So if we were to think of this graphically—if were to draw the graphs, if you shift the sine graph to the left by pi over 2 you get cosine then you get the cosine graph. And if you haven’t learned about shifting it, don’t worry about that or you might want to actually graph the two and I think you’ll get a sense of what I’m saying. So let’s do—I don’t know—and another way to rewrite this exact same thing is the sine of a is equal to the cosine of a minus pi over 2. If you just—let’s say I said that b is a plus pi over 2, let’s just say I said b is equal to a plus pi over 2 that we could say that this is b and then this would be b minus pi over 2. Remember switching around variables. I’m doing this in a much more loosey-goosey fashion that I normally do a lot of videos. But I want to show you that a lot of this Trigonometry can just be—it’s just kind of discovery. What’s the sine of a minus b? That looks a new one, doesn’t it? Well, let’s try to figure it out. Well, that equals sine of a cosine of minus b plus sine of minus b times the cosine of a. But what do we know about the cosine of minus b? We just—before I clear the screen we just figured it out that the cosine of minus b since it’s an even function is the same thing as the cosine of b. So we can rewrite that as, that equals to sine of a cosine of b. And then what’s the sine of minus b? Well, that’s the same thing as the minus sine of b, the last that we just prove that the sine of minus b that this is equal to minus sine of b. You could draw out the triangle and the unit circle if you don’t believe me but we just did that so that we can say that that is equal to minus sine of b cosine of a. Interesting. I encourage you do the same thing with the cosine of a minus b. These are all just—we’re using one or two or three trig identities together and trying to come up with new things. And I think at this point, we’ve literally got over everything that almost every trig identity you’ve seen in your book you can—you should be able to get there somehow just like keep on playing. And obviously all of these identities you can invert the sines and cosines and the tangents and you can get identities for secant and cotangent and cosecant and keep playing around. I encourage you to do so, do it graphically, draw the triangles. It’s also interesting to sometimes actually draw the graph on the x, y plain and so cosine of x plus pi over 2 or—because that sine of x plus pi over 2 or sine of x. And I think in the future, I’ll do a video where I really do explore all of that. Well, I hope I haven’t thoroughly confused you, I wanted to just kind of show you that a lot of trig, it’s really—it all comes from Socatoa and playing around with Socatoa and triangles and you can pretty much get—you can pretty much solve for everything you learned in Trigonometry. And if you don’t have Socatoa at least the units circle definition which actually better because this is more intensive. But anyway, that’s all for now. See you soon.

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Learn about Trig identities part 3 - part 5 if you watch the proofs

Welcome back. I have to— because I was hitting against the You Tube 10 minute limit. But I was always saying as we said cosine of minus a so I drew a right triangle with a and then I showed minus a and I said, “Well, all of the lengths are going to be the same but now the direction of—.” We’re kind of assuming this is all unit circles. If you don’t now remember unit circle, maybe you want to re-watch the videos we have on that.

But I’m just showing you that the cosine of minus a is equal to this side over the hypotenuse and this hypotenuse is the same as this hypotenuse. So cosine of minus a is adjacent over this hypotenuse but it’s the same thing so we know that cosine of minus a is equal to cosine of a. And actually by definition that makes it—I don’t want to confuse you too much but that makes cosine even functional. And I’ll show you more—actually I should do a whole presentation on even not functions.

And let’s see what sine of minus a is. Sine of minus a is equal to—so this is minus a, so it’s this side—so it’s the minus length of—let’s call this x, let’s call this y and let’s call this h. And if that is x, this is y, this length is y but this length right here is minus x. So the sine of minus a is minus x over h. What’s the sine of a? Sine of a is equal to—this is a opposite of hypotenuse x over h. So sine of minus a is equal to minus 1 times x over h or this is just the same thing at—I don’t know if we could multiply both sides of this by minus 1 minus x over h. So sine of minus a is equal minus sine of a.

So let me clear this out and rewrite this identity. And as you can see, all I’m doing is I’m just playing around with triangles and showing you that just using the basic Socatoa you can actually discover a whole set of Trigonometric identities. So let’s clear that and you might—it’s useful to memorize and normally advocate memorizing but it’s helpful just to do things quickly. But I’d also advocate being able to prove it to yourself so if you ever forget it and you don’t have a cheat-cheat available you can prove it and if you ever have to teach it then you’ll be I think able to explain the underlined themes a little bit better. So let’s clear this, let’s see if we can discover some more trig identities.

So we know that—so let’s see if we have sine, what’s the sine of a plus pi over 2? Well, we could use our handy sine of a plus b identity which we have already proved so we can use it now. So that’s the sine of a times the cosine of pi over 2 plus the sine of pi over 2 and we’re in radiance of course I could—this could have been 90 degrees instead if we wanted to be in degrees. Sine if pi over 2 times the cosine of a. Well, this equals to sine of a—what’s the cosine of pi over 2 or cosine of 90 degrees?

So that’s one. We’re on the unit circle where we’re pointing straight up so the x coordinate is zero, I could draw that out but I think you might want to draw the unit circle and figure out it for yourself or if you don’t—use a calculator but you will learn that it is zero, cosine of 5 over 2 is zero plus sine of pi over 2 for the same reason we’re pointing straight up on the unit circle so the y coordinate or the sine coordinate is 1 right on the unit—where it essentially is in the point zero 1 on the unit circle. So sine of pi over 2 is 1 so then times cosine of a. So sine of a times zero is zero, 1 times cosine of a is just cosine of a so we have a new useful trig identity. Sine of a plus pi over 2 is equal to cosine of a, fascinating.

So really, this is just telling us that cosine of a is the same thing as sine of a shifted. So if we were to think of this graphically—if were to draw the graphs, if you shift the sine graph to the left by pi over 2 you get cosine then you get the cosine graph. And if you haven’t learned about shifting it, don’t worry about that or you might want to actually graph the two and I think you’ll get a sense of what I’m saying. So let’s do—I don’t know—and another way to rewrite this exact same thing is the sine of a is equal to the cosine of a minus pi over 2.

If you just—let’s say I said that b is a plus pi over 2, let’s just say I said b is equal to a plus pi over 2 that we could say that this is b and then this would be b minus pi over 2. Remember switching around variables. I’m doing this in a much more loosey-goosey fashion that I normally do a lot of videos. But I want to show you that a lot of this Trigonometry can just be—it’s just kind of discovery.

What’s the sine of a minus b? That looks a new one, doesn’t it? Well, let’s try to figure it out. Well, that equals sine of a cosine of minus b plus sine of minus b times the cosine of a. But what do we know about the cosine of minus b? We just—before I clear the screen we just figured it out that the cosine of minus b since it’s an even function is the same thing as the cosine of b. So we can rewrite that as, that equals to sine of a cosine of b. And then what’s the sine of minus b?

Well, that’s the same thing as the minus sine of b, the last that we just prove that the sine of minus b that this is equal to minus sine of b. You could draw out the triangle and the unit circle if you don’t believe me but we just did that so that we can say that that is equal to minus sine of b cosine of a. Interesting.

I encourage you do the same thing with the cosine of a minus b. These are all just—we’re using one or two or three trig identities together and trying to come up with new things. And I think at this point, we’ve literally got over everything that almost every trig identity you’ve seen in your book you can—you should be able to get there somehow just like keep on playing.

And obviously all of these identities you can invert the sines and cosines and the tangents and you can get identities for secant and cotangent and cosecant and keep playing around. I encourage you to do so, do it graphically, draw the triangles. It’s also interesting to sometimes actually draw the graph on the x, y plain and so cosine of x plus pi over 2 or—because that sine of x plus pi over 2 or sine of x. And I think in the future, I’ll do a video where I really do explore all of that.

Well, I hope I haven’t thoroughly confused you, I wanted to just kind of show you that a lot of trig, it’s really—it all comes from Socatoa and playing around with Socatoa and triangles and you can pretty much get—you can pretty much solve for everything you learned in Trigonometry. And if you don’t have Socatoa at least the units circle definition which actually better because this is more intensive. But anyway, that’s all for now. See you soon.

Transcription by: Scribe4you Transcription Services