Find the remaining five trigonometric functions of θ.
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Question

Find the remaining five trigonometric functions of θ.

csc θ = -5/2, θ in quadrant III

Accepted Answer

Hey, everyone in this problem, we're asked to determine the exact values of the other five circular functions of theta. We're given that KC can of theta is equal to negative 11 divided by five and that theta is in quadrant four. OK. So when we're told, we wanna find the circular functions, we can also think of those as a trigonometric functions. OK. Those two words can be used interchangeably. So we're thinking about those trigonometric ratios, sine cosine tangent CC can and cotangent. OK. We already have cos all right. So let's start with what we were given. We were given KC can of theta. And let's recall in terms of our coordinates. OK. Koc can theta can be written as R divided by Y and we know that this is equal to negative 11 divided by five. All right. Now, we can use this what we have to define what R is and to define what Y is, but we have a negative. So we have to figure out where that negative goes. Does it go with the R? Does it go with the Y? OK. And in this case, there's two kind of ways to figure this out. Now R is always going to be po OK, remember that it measures a distance and so R is actually always going to be positive, which by default means that the negative must go with the Y. Now, the other way we can think about this is if we think about the quadrant that theta is in. So if we draw our Cartesian plane, we have the in quadrant four, we recall that quadrant four is the bottom right. OK. So quadrant four is below the X axis. So if we were to put a point in quadrant four and write the coordinates, the Y value would be negative. OK. So that checks that we're in quadrant four. So the Y value must be negative. And so R we can say is going to be equal to 11 and why is going to be equal to the negative five? And we're gonna put a red box around these because we're gonna need these values to determine those trigonometric ratios that we're looking for. Now there's one trigonometric ratio we can do right off the bat. That's a little bit simpler than the others in this case. And we want to think about the relationship between KC can of theta and sine of theta. They recall that KC can of theta is the inverse of sign the, which tells us that S data. And when I say inverse, I do not mean the mathematical inverse, I meant the reciprocal OK. It's the reciprocal one divided by s of data. All right. So that means that sine theta is the reciprocal of cosecant of data. OK? Sine theta is one divided by Kent of the we know Ko Si can't. So we can substitute this value and S is going to be one divided by negative 11, divided by five which gives us a value of negative five divided by 11. So this is the value of sine of theta. So that's one of those trigonometric ratios we were looking for. Now to figure out the other four, we're gonna need to do some calculations. OK. We have our R value, we have our Y value. The other value we need for all of these other ratios is X. OK. And recall that Xy and R are related to the Pythagorean theorem and they form a triangle, a right angle triangle. And so we can say that X squared plus Y squared is equal to R squared. We have Y, we have R we can solve for X substituting in our values X squared. What's negative five squared is equal to 11 squared. Simplifying X squared plus 25 is equal to 121. So X squared must be equal to 96 which gives us X equal to positive or negative square root of 96. Now, this is really, really important when we're doing trig problems. We have to remember that when we take the square root, we get both the positive and the negative root. OK. Now, based on the question, we want to infer which one makes sense. So let's go back to the little diagram. We drove a coordinate point, we're in quadrant four, that is to the right of the Y axis. So again, if we were to draw a point in quadrant four, the X coordinate would be positive. OK. So X is positive and coordinate four. So we are going to take the positive root X positive. All right. So we get that X is equal to positive the square root of 96. And we can simplify this. The square root of 96 can be written as the square root of 16 multiplied by six. We know that the square root of 16 is four. So we can write this as four multiplied by the square root of six. All right. So we figured out sign already the next trigger ratio we're gonna look at is cosine of the and recall that cosine of the is equal to X divided by R. OK. We have our X value now and we know our R value. And so cosine of data is gonna be four route six divided by 11. OK? Just substituting in those values. All right, let's move to the next one which is gonna be tangent of the and tangent of the recall is going to be equal to Y divided by X. Now, similarly tangent of data, we can write a sine of the divided by cosine of theta. We have sine and we have cosine. So we could use that as well. But because we have the Y and X values, we're just gonna substitute those indirectly, it's a little bit simpler, a little bit less simplification that needs to be done. All right, when we do this are Y value negative five divided by our X value for root six. Now we wanna go ahead and rationalize the denominator. We wanna get this square root out of the denominator. So we're gonna multiply the numerator and denominator by the square root of six. And we get that tangent of data is going to be negative five multiplied by the square root of six all divided by 24. All right. So we have tangent of data and now we've got to get to the other two functions. The first one being C can of theta recall that C can of theta is one divided by cosine of theta. We know cosine of the. So we're just going to reciprocate it. So we get 11 divided by four, route six. And just like with tangent, we want to rationalize that denominator multiplying by the square root of six in the numerator and denominator, we get 11 multiplied by route six divided by 24. And our final trig ratio is going to be cotangent. OK. So cotangent of theta is going to be equal to the reciprocal of tangent of theta. OK. Well, we know tangent of data and we're actually, instead of starting with the final version of tangent that we had, we're gonna start with the initial version of tangent we had before we rationalized because when we take the reciprocal that square root will be in the numerator, we don't have to worry about it. OK? All right. So when we do this, we're gonna get four multiplied by the square root of six divided by negative five. And those are all of the five trigonometric ratios we were looking for, we found sine right off the bat and then we have cosine tangent CC A and cotangent here. Thanks everyone for watching. I hope this video helped see you in the next one.

Find the remaining five trigonometric functions of θ.
csc θ = -5/2, θ in quadrant III

Key Concepts

Reciprocal Trigonometric Functions

Sign of Trigonometric Functions in Quadrants

Pythagorean Identity

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