SOLUTION: Problems containing reference angles/numbers. t = 9π/4 which as I understand it would go 9π/4 --> 2 1π/4 I follow the unit circle counterclockwise and pas

Algebra ->  Trigonometry-basics -> SOLUTION: Problems containing reference angles/numbers. t = 9π/4 which as I understand it would go 9π/4 --> 2 1π/4 I follow the unit circle counterclockwise and pas      Log On


   

Question 919518: Problems containing reference angles/numbers.
t = 9π/4
which as I understand it would go
9π/4 --> 2 1π/4
I follow the unit circle counterclockwise and pass 2π and go a little more for the quarter. Now since the 2π is closest between my calculations would then go
2π/1 - 9π/4 find a common denominator 8π/4 - 9π/4 = -π/4 is this the correct answer?
Problem 2: t = -10π/7 --> -1 3π/4
I follow the unit circle clockwise I go past -1π and then 3/4's to the middle of the fourth quadrant. Since it's nearing 2π it must therefore be 2π - 10π/7 find the common denominator it would then be
14π/7 - 10π/7 = 4π/7? is this one correct?
the other two are
t = 6
and
t = -7
which I have no clue how to solve.
Please help me
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Problems containing reference angles/numbers.
t = 9π/4=2π+π/4
rotate cc 2π + an aditional (1/4)π which terminates in quadrant I, making π/4 the reference angle in quadrant I.
..
t = -10π/7=-(π+3π/7)
rotate cw π + an aditional (3/7)π which terminates in quadrant II, making 3π/7 the reference angle in quadrant II.
..
t = 6(radians)
rotate cc 6 radians which terminates in quadrant IV just short of 2π(6.28) radians. The reference angle is then=6.28-6=.28 radians in quadrant IV.
..
t = -7(radians)
rotate cw 7 radians which terminates in quadrant IV. The reference angle is then=7-6.28=.72 radians in quadrant IV.
..
Note:The reference angle is the angle between the terminating side and the x-axis