Find the remaining five trigonometric functions of θ.
si...

Question

Find the remaining five trigonometric functions of θ.

sin θ = -4/5, cos θ < 0

Accepted Answer

Hey, everyone. Welcome back in this problem. We're asked to determine the exact values of the other five circular functions of theta. We're told that sine of theta is equal to negative 16 divided by 65. And that cosine of theta is less than zero. Now, when they say circular functions here, what they mean is trigonometric functions. OK? We can use those two words interchangeably. And so we're looking for those trigonometric ratios or given sine data, we wanna find cosine of data, tangent of data C can of the cosecant of data and cotangent of data. All right. So let's start with what we know. We've been told that SINE theta is negative 16 divided by 65. And recall that sine theta can be written as Y divided by R when we write up those coordinates. Yeah. And this is equal to negative 16 divided by 65. Now, our R value can't be negative. OK. Remember that the R value measures a distance that radius and so the R value is always positive. OK. So if sine is negative, that means that the negative must be associated with the Y value. OK. So we must have that Y is equal to negative 16 and R is equal to 65. So we have our Y value and we have our R value. Now, in order to determine all of these other functions, there's one more value that we need to calculate and that's the X value. But recall that Xy and R form a nice right triangle. So we can use the py figure in theorem. So we have that X squared plus Y squared is going to be equal to R squared. OK. And in this red box, I'm gonna put that Y value and R value we just found. So we don't lose track of it. We're not gonna need to keep coming back to it. And because we have those, we can use our Pythagorean theorem to solve for X substituting in our values X squared plus negative 16 squared is equal to 65 squared. Simplifying X squared plus 256 is equal to 4225. Subtracting by 256 on both sides, X squared is equal to 3969. And we can take the square root. And when we're working on trick problems like this, it's really important to remember that when we take the square root, we get both the positive and the negative root. OK. So we take that square root and we get positive or negative 63. OK. Now, once we have this, we need to infer from our problem, which makes sense. OK. So let's think about this. We're told that cosine of theta is less than zero. OK? We know that cosine of theta is equal to the X value divided by the R value R is positive, it's always positive. So if cosine is less than zero, it must mean that X is negative. OK? Just like when we had s being negative, we knew that the Y value is negative here, we're doing the same thing but with the X value, so we've taken the square root, we know mathematically we get the positive or negative root and then we need to infer from the question what our X value is and in this case, it's going to be negative 63 all right, red box around that one as well. So we have X, we have Y and we have R and those are the three things we need to write down all of the trigonometric ratios. Now we were given sin of the. So let's start with cosine of the cosine of theta is going to be equal to or we've just written it X divided by R. Our X value is negative 63 divided by RR value which is 65. So cosine of theta is going to be negative 63 divided by 65. Let's move to tangent of data. A tangent of data recall is going to be Y divided by X. Another way to think about this is we know that tangent is sine of the, divided by cosine of data. We now know sine and we know cosine. So we could use that instead as well. Um For this problem, we're gonna do the Y divided by X method. When we do that, we get negative 16 divided by negative 63 those negatives are gonna divide out and we're left with a positive value. So we get 16 divided by 63. OK. So that's our value for tangent of data. Now we can move on to the reciprocal function. So we have KC can of the recall that this is equal to one divided by sine of the, we were given sine theta in the problem as negative 16 divided by 65. And so the reciprocal of that is negative 65 divided by 16. OK. We have C can the next and recall C can theta one divided by cosine of theta. Hey, we know cosine of the, we just found it negative 63 divided by 65. And so when we reciprocate that we get negative 65 divided by 63 and our final trigonometric ratio is cotangent of theta. So cotangent of theta is going to be the reciprocal of tangent of the, we just found tangent of theta. We reciprocate it, we get 63 divided by 60 and so there we have it, we found all five of those other trigonometric ratios. So we have cosine of theta is negative 63 divided by 65 tangent of theta is 16 divided by 63 CO C can of theta is negative 65 divided by 16 C can of theta negative 65 divided by 63 and cotangent of theta. 63 divided by 16. Thanks everyone for watching. I hope this video helped see you in the next one.

Find the remaining five trigonometric functions of θ.
sin θ = -4/5, cos θ < 0

Key Concepts

Pythagorean Identity

Sign of Trigonometric Functions in Quadrants

Definitions of the Six Trigonometric Functions

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