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. (―8 , 15)

Accepted Answer

Hello, everyone. We're 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 rationalization. Wherever necessary, the point we're given is negative 912, we're provided a coordinate plane. And since the point negative 912 is a point on the terminal side, I'm gonna plot that burst. So negative nine and then up to 12 on my Y axis should be roughly here. So it's in quadrant two negative 912, I'm going to connect that line or that point with a line to the origin. And since we want it to be a positive angle from standard position, we're gonna start at the positive X axis and draw an angle or sorry, an arrow showing the rotation. And I'm gonna label that angle theta. So next, I need to find all trigonometric functions. So I recall that in order to find my trigonometric functions, I'm gonna start with some ratios and all of those ratios have XY or R. So X I can find from the point we were given. So X is negative nine Y is gonna be 12, R, I don't know what R is yet. So let's so for it. So I recall, I have a formula that R equals the square root of X squared plus Y squared. So that'll equal the square root of negative nine squared plus 12 squared. So when I do those squares out, I get the square root of 81 plus 144 and that is the square root of 225 and the squared of 225 is 15. So R is going to be 15. So now we can go through and do all of our functions. So I'm going to start with sign the sign of the is Y divided by R. So why is 12, R is 15? So I have 12/15 but I do wanna reduce that. And since it can be reduced by three, we will get 4/5. So the sin of theta is 4/5 next, the co sin theta and that is X divided by R. So X is negative nine, our is 15, again, negative 9/15 can be reduced by three. So we'll get that the cosine of theta is negative 3/5. The tangent of data is Y divided by X. So Y again was 12, X is negative nine. We wanna simplify that. So again, I'm reducing by three and I get negative four thirds. So the tangent of data is negative four thirds. Next, I'm gonna do the Koi Can of theta and that ratio is R divided by Why? So that would be 15 divided by 12 which when reduced is five fourths. Next, the sea can of theta, it will be R divided by X. So that would be 15 divided by negative nine which again will reduce, we'll get negative five thirds. So that is the C can of data. And our last function is the cotangent of the and that's going to be X divided by Y. So negative nine divided by 12 which I will reduce to negative 3/4. So the cotangent of data is negative 3/4. So that is all six of them. We have found all of them and plotted the point. So we are all set, 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. (―8 , 15)

Key Concepts

Standard Position of an Angle

Finding the Reference Angle and Least Positive Angle

Six Trigonometric Functions from a Point on the Terminal Side

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