SOLUTION: This is a test preparation question but I just am not sure how to solve it or where to start: If sin t = 1/3 and pi/2 < t < pi, then determine the exact value of cos t, sin(2pi

Algebra ->  Trigonometry-basics -> SOLUTION: This is a test preparation question but I just am not sure how to solve it or where to start: If sin t = 1/3 and pi/2 < t < pi, then determine the exact value of cos t, sin(2pi       Log On


   

Question 737601: This is a test preparation question but I just am not sure how to solve it or where to start:
If sin t = 1/3 and pi/2 < t < pi, then determine the exact value of cos t, sin(2pi - t), and cos(2pi - t).
the answer is -2sqrt%282%29%2F3,+-1%2F3,+-2sqrt%282%29%2F3
Thank you in advance!
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
If sin t = 1/3 and pi/2 < t < pi, then determine the exact value of cos t, sin(2pi - t), and cos(2pi - t).
***
let O=opposite side
let A=adjacent side
let H=hypotenuse
...
sin t=1/3=O/H
O=1, H=3
A=sqrt%28H%5E2-1%5E2%29=sqrt%289-1%29=-sqrt%288%29
cos%28t%29=A%2FH=-sqrt%288%29%2F3=-2sqrt%282%29%2F3
...
sin%282pi-t%29=-1%2F3 (Note: this moves the standard position of t from Q2 to Q3 where sin<0, with the same reference angle) You could use the sin addition formula to show this.
...
cos%282pi-t%29=-2sqrt%282%29%2F3(Note: this moves the standard position of t from Q2 to Q3 where cos<0, with the same reference angle) You could use the cos addition formula to show this.