## Need some clarification on terminal points

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.
agenta1013
Posts: 7
Joined: Thu Oct 17, 2013 2:20 pm
Contact:

### Need some clarification on terminal points

I've gotten a few problems like these on homework and quizzes, and yet again no class or textbook demonstrations. Could someone help point me in the right direction for solving these?

1. Terminal point t is at point $(3/5,4/5)$ Compute the following:

$pi-t = (-3/5, 4/5)$

$-t = (3/5, -4/5)$

$pi+t = (-3/5, -4/5)$

$2pi+t = (3/5,4/5)$

If x and y are both positive, then:

$pi-t = (-x,y)$

$-t = (x, -y)$

$pi+t = (-x, -y)$

$2pi+t = (x,y)$

Any hints on how these change if x or y start out negative?

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
1. Terminal point t is at point $(3/5,4/5)$ Compute the following:

$pi-t = (-3/5, 4/5)$

$-t = (3/5, -4/5)$

$pi+t = (-3/5, -4/5)$

$2pi+t = (3/5,4/5)$
Are the values after the "equals" signs meant to be the answers, so the "questions" were the parts before the "equals" signs? Is "t" a point or an angle measure? If the former, then expressions such as "$\pi\, -\, t$" have no meaning, since they represent an ordered pair being subtracted from a single numerical value (or else an angle measure, stated in radians). If the latter, then perhaps the original point is meant to be a representative point lying on the terminal side of the underlying angle. So maybe you're supposed to find the "corresponding" point for the listed angle values?

(By the way, if this exercise is representative of the "clarity" of your textbook, then it's no wonder you're a bit confused!)

If x and y are both positive, then:

$pi-t = (-x,y)$

$-t = (x, -y)$

$pi+t = (-x, -y)$

$2pi+t = (x,y)$
I think the basic idea is to move the point from quadrant to quadrant, based on the amount of change. Pi is halfway around the circle; 2pi is all the way 'round. So "pi - t" is "start at the positive x-axis, go around halfway to the negative x-axis, and then subtract backwards (toward Quadrant I) as far as the original terminal side had been from the positive x-axis". Similarly, "2pi + t" is "go all the way around, so you're right back where you started, and go as far around as the terminal side of the original angle, so you're right where the terminal side of t had been".

For x and y both positive, the terminal side of t (I'm assuming they meant t to be an angle measure, not a point) has to be in Quadrant I. As a result, -t must be in Quadrant IV, pi + t must be in Quadrant III, and pi - t must be in Quadrant II. So assign the signs as appropriate to the quadrant.
Any hints on how these change if x or y start out negative?
If they've both negative, then the point is in Quadrant III. Then -t must be in Quadrant II, pi - t must be in Quadrant IV, pi + t must be in Quadrant I, and 2pi + t must be in Quadrant III. Draw a picture, if you're not sure.