Sketch an angle θ in standard position such that θ has the lea...

Question

Sketch an angle θ in standard position such that θ has the least positive measure, and the given point is on the terminal side of θ. Then find the values of the six trigonometric functions for each angle. Rationalize denominators when applicable. See Examples 1, 2, and 4. (―12 , ―5)

Accepted Answer

Hello, everyone. We are asked to plot the least positive measure of the given angle in standard position. Such that the given point is on the terminal side. Also, we want to write the value of all trigonometric functions for the given angle performing rationalizations. When necessary, we're given the point negative eight negative six, we have a blank coordinate plane with no values marked. So the first thing I would like to do is plot the point negative eight negative six. It might not be precise, but I wanna find generally where that lies. So it's negative eight on the X and negative six on the Y. So that would fall in quadrant three. So if this is the terminal side, I'm gonna draw a line connecting that point to the origin. And if we're in standard form or sorry standard position, the beginning of our angle will be on the positive side of the X axis. So our angle goes from the positive side of the X axis all the way around into quadrant three. And I'm gonna call that theta just so we can name the angle. So to find all six trig functions of data, we need to see what are we given? What do we need to find? Well, because we're given a point, we can think about the ratios we know for our trig functions X is gonna be negative eight. Why is negative six? And recall to do these ratios, we need to find the value of R R is like our hypotenuse. So we will use the Pythagorean theorem to find the value of R so X squared. So eight negative eight squared plus Y squared. So negative six squared equals R squared. So 48 plus 36 equals R squared would be 100 equals R squared. And when I take the square root of both sides, R will equal 10 and R is positive because R is like our radius and a radius is a distance and distances are always positive. So now that we know Xy and R, we can start finding all of our trick functions. So starting with the sign of theater, recalling that sine is the ratio of Y divided by R Y is negative six divided by 10 which will reduce to negative 3/5. Next cosine of the, this ratio is X divided by R X is negative eight divided by R which is 10. And when I reduce that I get negative 4/ the tang of the is the ratio of Y divided by X. So Y is negative six divided by X which is negative eight. And when I reduced that I would get 3/4. So three down three to go the sea camp of theta. Recall that sea can is like the reciprocal of cosine. So this would be R divided by X. So this would be 10 divided by negative eight which reduces to negative five fourths. Coy can of theta. Again, recall co C can is the reciprocal of sine. So this would be R divided by Y so divided by negative six which when reduced is negative five thirds and the cotangent at theta will be the reciprocal of the tangent. So this would be X divided by Y. So negative eight divided by negative six which reduces to four thirds. So we have the sign of theta is negative 3/5 cosine of theta is negative 4/5 tangent of theta is 3/4 C can of theta is negative five fourths co CK of theta is negative five thirds. Cotangent of theta is four thirds. Have a nice day.

Sketch an angle θ in standard position such that θ has the least positive measure, and the given point is on the terminal side of θ. Then find the values of the six trigonometric functions for each angle. Rationalize denominators when applicable. See Examples 1, 2, and 4. (―12 , ―5)

Key Concepts

Standard Position of an Angle

Finding the Reference Angle and Least Positive Angle

Six Trigonometric Functions from Coordinates

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