Algebra/Trig Review - Pauls Online Math Notes - Lamar University

Algebra/Trig Reviewdetermine what this angle is. When we look for these angles we typically wantpositive angles that lie between 0 and 2π . This angle will not be the only possibilityof course, but by convention we typically look for angles that meet these conditions.To find this angle for this problem all we need to do is use a little geometry. Theangle in the first quadrant makes an angle of 6π with the positive x-axis, then so mustπthe angle in the fourth quadrant. So we could use − , but again, it’s more common6π 11πto use positive angles so, we’ll use t = 2π− = .6 6We aren’t done with this problem. As the discussion about finding the second anglehas shown there are many ways to write any given angle on the unit circle.πSometimes it will be − that we want for the solution and sometimes we will want6both (or neither) of the listed angles. Therefore, since there isn’t anything in thisproblem (contrast this with the next problem) to tell us which is the correct solutionwe will need to list ALL possible solutions.This is very easy to do. Go back to my introduction in the Trig Function Evaluationsection and you’ll see there that I usedπ+ 2 π n, n= 0, ± 1, ± 2, ± 3, 6to represent all the possible angles that can end at the same location on the unit circle,i.e. angles that end at 6π . Remember that all this says is that we start at 6π then rotatearound in the counter-clockwise direction (n is positive) or clockwise direction (n isnegative) for n complete rotations. The same thing can be done for the secondsolution.So, all together the complete solution to this problem isπ+ 2 π n, n= 0, ± 1, ± 2, ± 3, 611π+ 2 π n, n= 0, ± 1, ± 2, ± 3, 6πAs a final thought, notice that we can get − by using n = − 1 in the second solution.62. 2cos( ) 3Solutiont = on [ − 2 π,2 π]© 2006 Paul Dawkins 63http://tutorial.math.lamar.edu/terms.aspx