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Trigonometry - Example 4

Trigonometry is the branch of mathematics concerning the relationships between the sides and the angles of triangles. The field emerged in the Hellenistic world during the 3rd century BC from applications of geometry to astronomical studies. Trigonometry is also the foundation of surveying.


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Video Transcript

welcome to the last Trigonometry Review video, and this one will consider functions are rather equations that include trig functions. Eso one of the most common ones that we will come across later in the course Looks something like this. Position is a function of time is equal to a cosign of omega T plus five. Don't let the new names to the different variables throw u. K A. Is still gonna be the amplitude as it was before. In this case, though, Omega is going to this kind of curly w looking thing. It's a Greek letter. Omega is going to be equal to our frequency or in this case, we call it angular frequency. And then Fi is gonna be equal to our phase shift or, more generally, called the phase constant in physics. Okay, so say were given an equation like this and we're at We're saying, Hey, at what time? We have a particular position we wanna be able to solve for something inside the cosine function here. So we need to be able to do that algebra quickly. So say were given a particular position x not. And it's equal to a cosign of Omega t not plus fi. Okay, I want to solve for teen up here, divide over a and then used the inverse cosine function so that will look like a cozy into the negative one of X, not over a is equal to Omega Times, Tina plus Fi. And that was pretty easy. We subtract if I and then divide by Omega and were able to come up with an answer for what Tina ought to be. Okay. And it's by divided by Omega. And we found Tina. Okay, so sometimes it can be a fairly simple way to solve things just by brute force. Inverse functions. Other times, it's not quite so easy to come across an answer. For example, sometimes when you're dealing with inclined planes, you might have one equation that it looked like a times sign of X is equal to, um, we'll just put another constant air B, and then down here will have C times. Cosine of X is equal to D for a B, and C and D are all some Constance okay. And I want to solve for what, two of these variables, our se x and D. This could be a very difficult problem to solve because we have different trig functions here. You can go around and around all day, but there's a really simple trick. And that is if you divide the two functions by each other. Okay, so notice now we have a oversee time. Sign of X over CoSine X will sign over co sign that's equal to Tangent of X is equal to be over D. Okay, so now that we have this, if we're clever about how we multiply things what multiplication, What multiplier we put in front of these two functions. Say, if I were to multiply both sides of this by three or something like this, we can actually cause one of our unknowns to disappear and very quickly solve for X. So I could have Tangent of X is equal to see over a times B over D and I can solve for X using the inverse tangent function. Okay, um so you'll see that I will do this example twice in the unit where it finally arises. So no need to worry about it now. If you're not sure, you understand. But I want to point out to you that this is a legitimate way of solving. If you have a sign and a cosign and two different equations that are supposed to be related to each other somehow, then you're able to combine them by dividing one the sign over the co sign and then the equivalent sides over each other. And you can come up with tangent here and make your life a lot easier. Um, the other thing to remember that triggered a metric functions have things called trig and metric identities that you may or may not have learned. One of the most common ones looks like there's science squared X plus cosine squared X equals one. If you don't believe me, type it into your calculator and take a look at it promised equals one. Just type in triggered a metric identities into the Internet. You'll find list upon list, and you can use those to solve for these functions as well