Learn about Using Trig Functions Video

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Learn about Using Trig Functions

Learn about Using Trig Functions We’re now going to do a few examples to actually show you why the trig functions are actually useful. So let’s get started with the problem. Let’s say I have this right triangle, that’s my right triangle and there’s a right angle. And let say I know that the measure of this angle is pi over 4 radiance and I’ll just rad for short. If the measure of this angle is p0i over 4 radiance and I also know that this side of the triangle, this side right here is tan square roots of 2. So now I know this side of the triangle, I know this angle which is pi over 4 radiance and now the question is, what is this side of the triangle? I’m going to highlight that and let me make it in orange. So let’s figure out what we know and what we need to figure out. We know the angle pi over 4 radiance and actually it turns out if you were to convert that to degrees, it would be 45 degrees. And we know what side is this, this is the hypotenuse of the triangle. And what are we trying to figure out? Are we trying to figure out the hypotenuse, the adjacent, side to the angle or the opposite side of the angle? Well, this is the hypotenuse we already know that. This is the opposite side and this yellow side is the adjacent side. It’s just adjacent to this angle. So we know the angle, we know the hypotenuse and we want to figure out the adjacent side. So let me ask you a question. What trig function deals with the adjacent side and the hypotenuse because we have adjacent side what we want to figure out and we know the hypotenuse. Well, let’s write down a mnemonic just incase you forgot it, Socatoa. So which one uses adjacent then hypotenuse, it’s ca. And ca—the c is for what? The c is for cosine, cosine of a angle let’s just call any angle is equal to the adjacent of the hypotenuse. So let’s use this information to try to solve for this—orange side or this yellow side. So we know that cosine of pi over 4 radiance so let’s say cosine of pi over 4 must equal this adjacent side right here—let’s just call that a, a for adjacent. The adjacent side divided by the hypotenuse, the hypotenuse is this side and the problem we were given that it’s tan square roots of 2. So we can solve for a by multiplying both sides of this equation by tan square roots of 2 and we will get—because right if you just we multiply times tan square root of 2, this cancel out and you get a tan square roots of 2 here. So you get a is equal to tan square roots of 2 times the cosine pi over 4. Now, you probably saying, Sal, this doesn’t look too simple and I don’t know how big the cosine of pi over 4 is, what do I do? Well, no one has the trig functions or the values of the trig function memorized. There is couple of ways to do it. Either I could give you what the cosine of pi over 4 is and that’s sometimes given in a problem or you can make sure that your calculator is set to radiance and you could just type in pi divided by 4 which is roughly .79 and then press the cosine button, you find now what is good for and we’ll get a value. Or and this is kind of the old school way of doing it, there are trig tables where you could look up what cosine of pi over 4 is in the table. Since I don’t have any of that at my disposal right now, I’ll just tell you what the cosine of pi over 4 is. The cosine of pi over 4 is square root of 2/2. So a, which is the adjacent side, a for adjacent, is equal to tan square roots of 2 times square root of 2/2. Remember, to get the square root of 2/2 you might be a little confuse you’re like how did Sal get that? All I said is, the cosine of pi over 4 is square root of 2/2 and that’s not something that—well, actually this when you might not know often because it’s the 45 degree angle. But this isn’t something people memorize, this is something you’d look up it’s given in the problem or you’d use a calculator for it. And then the calculator of course wouldn’t give you square root of 2/2, it will give you a decimal number that’s not obvious the square root of 2/2. But anyway, I told you that the cosine of pi over 4 is the square root of 2/2 and so if we multiply--what’s the square root of 2/2, what’s the square root of 2 times square root of 2, it’s 2. So that’s 2 and that cancels with that 2 and so everything cancels except for the tan so the adjacent side is equal to tan. Let’s do another one. Let me delete this, give me one second. I’m actually—this is one the fewer modules that I’m not generating the problems on the fly because I need to make sure that I actually have the trig function I’ll use before I do the problem. So let’s say, I have another right triangle, another right—actually I probably should have deleted that last one. So let’s say this is my right triangle--how much time do I have? Four minutes left. So this is my right triangle and I know the angle—I know this angle right here is .54 radiance and I also know that this side right here is 3 units long and I want to figure out this side. So what I do know? Well, this side is what side relative to the angle; it’s the opposite side because the angle is here we go opposite the angle so this is the opposite side. And what’s this side? Is this the adjacent side or is it the hypotenuse? Well, this is the hypotenuse, the long side and this opposite the right angle. So this is the adjacent side. So what trig function uses opposite and adjacent? Let’s write down Socatoa again. Toa uses opposite and adjacent oa so t for tangent, toa. So tangent is equal to opposite over adjacent. So let’s use that. So let’s take the tangent of .54 radiance so the tangent of .54 will equal the side opposite to it so that’s 3, the opposite side of 3 over the adjacent side. Once again, the adjacent side is what we don’t know so we have to solve for a. So if we multiply both sides by a, we get the tan of .54 and we could do that because we know it’s not zero equals 3 or a is equal to 3 divided by the tangent of .54. So once again, I don’t have memorized what the tangent of .54 but I will tell you what it is because you also don’t have to memorize or you could a calculator if you had radiant function. The tangent of .54 is equal to—let me make sure I have this right—the tangent of .54 is 3/5 so then a is equal to 3/5. The adjacent side—and once again, how did I get this 3/5? Well, I just told you or you could use a calculator know that the tangent of .54 is 3/5 and of course I’m using numbers that work out well just so that the fractions all cancel. But—so we know the adjacent side is equal to when you divided by fractions like multiplying by the numerator—multiplying by the inverse so times 5 over 3, so the adjacent side is equal to 5. There you go. So let’s just think about what I always do. I think about what I have, what sides I have and what side I want to solve for in this case, it was the opposite side I had and I want to solve for the adjacent side and I said, what trig function involves those two sides the opposite and the adjacent? I wrote down Socatoa and I said, toa, opposite and adjacent that’s tan. So I took the tan of the angle and then I said the tan of the angle is equal to the opposite side divided the adjacent side, that’s right here, and I just solved for the adjacent side and of course I used the calculator or I told you with the tangent of .54 is. I think I’ll do a couple more of this problem in the next module but I’m out of time for now. Have fun.

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Learn about Using Trig Functions

We’re now going to do a few examples to actually show you why the trig functions are actually useful. So let’s get started with the problem. Let’s say I have this right triangle, that’s my right triangle and there’s a right angle. And let say I know that the measure of this angle is pi over 4 radiance and I’ll just rad for short. If the measure of this angle is p0i over 4 radiance and I also know that this side of the triangle, this side right here is tan square roots of 2. So now I know this side of the triangle, I know this angle which is pi over 4 radiance and now the question is, what is this side of the triangle? I’m going to highlight that and let me make it in orange.

So let’s figure out what we know and what we need to figure out. We know the angle pi over 4 radiance and actually it turns out if you were to convert that to degrees, it would be 45 degrees. And we know what side is this, this is the hypotenuse of the triangle. And what are we trying to figure out? Are we trying to figure out the hypotenuse, the adjacent, side to the angle or the opposite side of the angle? Well, this is the hypotenuse we already know that. This is the opposite side and this yellow side is the adjacent side. It’s just adjacent to this angle. So we know the angle, we know the hypotenuse and we want to figure out the adjacent side. So let me ask you a question. What trig function deals with the adjacent side and the hypotenuse because we have adjacent side what we want to figure out and we know the hypotenuse. Well, let’s write down a mnemonic just incase you forgot it, Socatoa.

So which one uses adjacent then hypotenuse, it’s ca. And ca—the c is for what? The c is for cosine, cosine of a angle let’s just call any angle is equal to the adjacent of the hypotenuse. So let’s use this information to try to solve for this—orange side or this yellow side. So we know that cosine of pi over 4 radiance so let’s say cosine of pi over 4 must equal this adjacent side right here—let’s just call that a, a for adjacent. The adjacent side divided by the hypotenuse, the hypotenuse is this side and the problem we were given that it’s tan square roots of 2. So we can solve for a by multiplying both sides of this equation by tan square roots of 2 and we will get—because right if you just we multiply times tan square root of 2, this cancel out and you get a tan square roots of 2 here.

So you get a is equal to tan square roots of 2 times the cosine pi over 4. Now, you probably saying, Sal, this doesn’t look too simple and I don’t know how big the cosine of pi over 4 is, what do I do? Well, no one has the trig functions or the values of the trig function memorized. There is couple of ways to do it. Either I could give you what the cosine of pi over 4 is and that’s sometimes given in a problem or you can make sure that your calculator is set to radiance and you could just type in pi divided by 4 which is roughly .79 and then press the cosine button, you find now what is good for and we’ll get a value. Or and this is kind of the old school way of doing it, there are trig tables where you could look up what cosine of pi over 4 is in the table. Since I don’t have any of that at my disposal right now, I’ll just tell you what the cosine of pi over 4 is.

The cosine of pi over 4 is square root of 2/2. So a, which is the adjacent side, a for adjacent, is equal to tan square roots of 2 times square root of 2/2. Remember, to get the square root of 2/2 you might be a little confuse you’re like how did Sal get that? All I said is, the cosine of pi over 4 is square root of 2/2 and that’s not something that—well, actually this when you might not know often because it’s the 45 degree angle. But this isn’t something people memorize, this is something you’d look up it’s given in the problem or you’d use a calculator for it. And then the calculator of course wouldn’t give you square root of 2/2, it will give you a decimal number that’s not obvious the square root of 2/2. But anyway, I told you that the cosine of pi over 4 is the square root of 2/2 and so if we multiply--what’s the square root of 2/2, what’s the square root of 2 times square root of 2, it’s 2. So that’s 2 and that cancels with that 2 and so everything cancels except for the tan so the adjacent side is equal to tan.

Let’s do another one. Let me delete this, give me one second. I’m actually—this is one the fewer modules that I’m not generating the problems on the fly because I need to make sure that I actually have the trig function I’ll use before I do the problem. So let’s say, I have another right triangle, another right—actually I probably should have deleted that last one. So let’s say this is my right triangle--how much time do I have? Four minutes left. So this is my right triangle and I know the angle—I know this angle right here is .54 radiance and I also know that this side right here is 3 units long and I want to figure out this side. So what I do know? Well, this side is what side relative to the angle; it’s the opposite side because the angle is here we go opposite the angle so this is the opposite side. And what’s this side? Is this the adjacent side or is it the hypotenuse? Well, this is the hypotenuse, the long side and this opposite the right angle. So this is the adjacent side. So what trig function uses opposite and adjacent? Let’s write down Socatoa again. Toa uses opposite and adjacent oa so t for tangent, toa. So tangent is equal to opposite over adjacent. So let’s use that.

So let’s take the tangent of .54 radiance so the tangent of .54 will equal the side opposite to it so that’s 3, the opposite side of 3 over the adjacent side. Once again, the adjacent side is what we don’t know so we have to solve for a. So if we multiply both sides by a, we get the tan of .54 and we could do that because we know it’s not zero equals 3 or a is equal to 3 divided by the tangent of .54. So once again, I don’t have memorized what the tangent of .54 but I will tell you what it is because you also don’t have to memorize or you could a calculator if you had radiant function. The tangent of .54 is equal to—let me make sure I have this right—the tangent of .54 is 3/5 so then a is equal to 3/5. The adjacent side—and once again, how did I get this 3/5? Well, I just told you or you could use a calculator know that the tangent of .54 is 3/5 and of course I’m using numbers that work out well just so that the fractions all cancel. But—so we know the adjacent side is equal to when you divided by fractions like multiplying by the numerator—multiplying by the inverse so times 5 over 3, so the adjacent side is equal to 5. There you go.

So let’s just think about what I always do. I think about what I have, what sides I have and what side I want to solve for in this case, it was the opposite side I had and I want to solve for the adjacent side and I said, what trig function involves those two sides the opposite and the adjacent? I wrote down Socatoa and I said, toa, opposite and adjacent that’s tan. So I took the tan of the angle and then I said the tan of the angle is equal to the opposite side divided the adjacent side, that’s right here, and I just solved for the adjacent side and of course I used the calculator or I told you with the tangent of .54 is. I think I’ll do a couple more of this problem in the next module but I’m out of time for now. Have fun.

Transcription by: Scribe4you Transcription Services