Give all six trigonometric function values for each angle θ. Rati...

Question

Give all six trigonometric function values for each angle θ. Rationalize denominators when applicable. See Examples 5–7.   sin θ = √5/7 , and θ is in quadrant I .

Accepted Answer

Welcome back. I am so glad you're here. We're asked to enumerate the value of the other five trigonometric functions of P given the following trigonometric function and condition rationalize the denominators if necessary, the trigonometric function that we're given is the sign of P which is equal to in the numerator, the square of 10 divided by eight. And we're told that P is in quadrant one. All right. First of all, what are those other five trigonometric functions of P? Well, those are going to be the cosine of P, the tangent of P, the cotangent of P, the secant of P in the Koi Kant of P. And what else do we know how can we get from the sign to those other five? We recall from previous lessons that the sign of an angle is equal to Y divided by R. So we know that our Y is equal to what's in the numerator? That's the square root of 10 and our R that's equal to what's in the denominator. That's eight. For these other five trigonometric functions, we know that the cosine of an angle is equal to X divided by R, the tangent of an angle is equal to Y divided by X. The cotangent of an angle is X divided by Y. The C can of an angle is R divided by X and the cos of an angle is R divided by Y. So we know our, our Y and our R but we don't know our X yet, but we can figure that out. This one takes a little bit of work but we can do it. We know that R is equal to the square root of X squared plus Y squared. And so we know our R that's eight equals the square at of, we don't know X squared yet plus Y. We know that's the squared of 10 squared. Now, we can simplify eight equals the square of X squared plus the squared of 10 squared is 10. Now, we wanna get that X by itself. So we're going to need to square both sides to get it out of the radical on the left eight squared is 64 that equals on the right, the square root of X squared plus 10 squared is just X squared plus 10. Now, we can subtract 10 from both sides on the left. 64 minus 10 is 54 that equals on the right, just X squared, we can take the square root of both sides. And so we've got an X equal to the square of 54 but we can simplify that I'm going to be simplifying radicals pretty quickly if you need a refresher on simplifying radicals, definitely go check out previous lesson videos. So for the square of 54 that's the same as the square of nine multiplied by six. And we know that the square of nine is three. So this simplifies 23 square at six. And so now we can plug in for our Xs and Ys and R's. But we want to keep one really important thing in mind here. We're told that P is in quadrant one. And that's relevant. If we draw a rough sketch of a coordinate plane, we've got a vertical Y axis and a horizontal X axis. They come together at the origin in the middle, we have four quadrants up into the right of the origin is quadrant one, up into the left of the origin is quadrant two, down into the left of the origin is quadrant three and down into the right of the origin is quadrant four. We're told that P is in quadrant one. What do we know about whether P is going to be positive or negative in quadrant one? We recall from previous lessons, we have a pneumonic device to help us remember where the different trigonometric functions are positive and that pneumonic device is all. So we've got an A in quadrant one students, we've got an S in quadrant two take, we've got A T in quadrant three, calculus. We've got ac in quadrant four and the, all from all or the A from all in quadrant one still stands for all when it comes to the trigonometric functions, because they are all positive in quadrant one. So that means when we plug in our X's Ys and R's our X values and our Y values, they're all positive as we're plugging them in. So we can now proceed. All right. So for the cosine, we know that's X divided by R our X in the numerator is three square root six R our R in the denominator is eight. That one is simplified. And that is the simplest one. We're going to do the rest of these get to be a little bit more fun. So we've got the cosine. Now, for the tangent, we know the tangent is Y divided by XY in the numerator. That's the square out of 10 divided by X in the denominator. That's three square at six. Here is where we get to rationalize the denominators. If necessary, we want that radical, the square out of six out of the denominator. So we multiply both the numerator and the denominator by the square out of six. And again, like I said, with simplifying Xon or simplifying our radicals, if you need a refresher on rationalizing denominators, definitely check out previous lesson videos because I'm going to go kind of fast in the numerator. The square root of 10 multiplied by the squared of six is the square rot of in the denominator, three square rot, six multiplied by the square rot of six is three multiplied by six. In the numerator, the square rot of 60 is the same as the square root of six multiplied by 10. But six and 10 both have a two as a common factor. So we can take a two out because we've got factors in there. And then what's left is a three multiplied by a five, which is 15. So inside we've got two square at 15, we could have also thought of the square out of 60 as four multiplied by 15. Well, the square root of four multiplied by 15. All right, in the denominator, three multiplied by 16. 3, multiplied by six is 18. And then one last bit of simplification, two and 18 are both divisible by two. So, in the numerator, we've got the square root of 15 that's divided by, in the denominator, we have nine. All right. So just three to go for our cotangent, we've got X divided by YX is three square rot six divided by Y which is square at 10. We need to rationalize. So we'll multiply both the numerator and the denominator by the square root of 10. In the numerator, we have three square foot six multiplied by the square rot of 10, which is three square foot 60 divided by square at 10, multiplied by square root. 10 is just 10 in the denominator And so a couple of things to simplify, we know from the previous one, that square rot of 60 is two square rot 15. So three multiplied by two is going to be six. So we've got six square at 15 divided by 10, both the six and the 10 are divisible by two. So this simplifies to be in the numerator three square at 15 divided by five. All right. So C can we have R divided by Xr is eight, divided by X is three square at six. Need to rationalize, we'll multiply both the numerator and the denominator by the square of six. And so in the numerator, we have eight square foot six. Let's divide it by, we know when we rationalize that denominator that's going to be 18 from the tangent eight and 18 are both divisible by two. So, simplifying here in the numerator, we've got four square foot six that's divided by nine. All right. Last one, we've got R divided by yr is eight in the numerator divided by Y is the square root of 10. Need to rationalize this one. So we're multiplying both the numerator and the denominator by the square rot of 10. In the numerator, we've got eight multiplied by the square root of 10, which is just eight square rot 10 divided by square rot of 10, multiplied by itself is just 10. Both eight and 10 are divisible by two. So this one will simplify to B in the numerator four square at 10 divided by five. And there are our five other trigonometric functions of P. So the cosine of P equals three square watt six divided by eight. The tangent of P equals the square root of 15, divided by nine. The cotangent of P equals three square ro 15 divided by five. The CK of P equals four square root six divided by nine. And the CO C can of P equals four square at 10, divided by five. Well done. We'll catch you on the next one.

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Give all six trigonometric function values for each angle θ. Rationalize denominators when applicable. See Examples 5–7. sin θ = √5/7 , and θ is in quadrant I .

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