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

Algebra/Trig Review3Now, there are no angles in the first quadrant for which sine has a value of − .2However, there are two angles in the lower half of the unit circle for which sine will3have a value of − . So, what are these angles? A quick way of doing this is to, for23a second, ignore the “-” in the problem and solve sin ( x ) = in the first quadrant2πonly. Doing this give a solution of x = . Now, again using some geometry, this3tells us that the angle in the third quadrant will be 3π below the negative x-axis orπ 4πππ + = . Likewise, the angle in the fourth quadrant will below the positive x-3 33π 5πaxis or 2π − = .3 3Now we come to the very important difference between this problem and Problem 1.The solution is NOT4πx= + 2 π n, n= 0, ± 1, ± 2, 35πx= + 2 π n, n= 0, ± 1, ± 2, 3© 2006 Paul Dawkins 65http://tutorial.math.lamar.edu/terms.aspx