## mattlai7 2 years ago if cos(t)=-1/5 and pi<t<3pi/2. find csc(t)

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1. zepdrix

|dw:1362772905106:dw|Understand why I drew that triangle in the third quadrant?

2. mattlai7

not quite sure why?

3. zepdrix

|dw:1362773055192:dw| They told us that the angle t in within this interval.$$\large \pi<t<3\pi/2$$. So we draw a line in the third quadrant and form a triangle from it.

4. mattlai7

ok now i understand that part

5. zepdrix

|dw:1362773145892:dw|Let's look at just the triangle now. Don't worry about the fact that I labeled the angle $$\large t'$$. That's not super important. Just think of it as $$\large t$$.

6. zepdrix

$$\large \cos t=-\dfrac{1}{5}=\dfrac{adjacent}{hypotenuse}$$ So how would we label these sides? HINT: We always let the hypotenuse be positive.

7. mattlai7

|dw:1362773386830:dw|

8. zepdrix

Ok looks good! Just need to find the missing side using the Pythagorean Theorem and then we just have one small step after that.

9. mattlai7

opposite will equal 4?

10. mattlai7

sorry did an error......missing side equals to $+/- 2\sqrt{6}$

11. zepdrix

Ok looks good. Now remember that we drew our triangle in the third quadrant. So for that OPPOSITE leg, did we move UP or DOWN? Do we want the positive or negative answer?

12. mattlai7

we want a negative answer because tan is only positive in the 3rd Q

13. zepdrix

Haha that's an interesting way to think about it XD I was thinking about the fact that we moved downward in the y direction to make the opposite length. So it would be negative. But whatever works for you! :D|dw:1362773803586:dw|

14. zepdrix

Ok good. Everything is labeled correctly. Just have to find $$\large \csc t$$ now! :)

15. mattlai7

hypotenuse over opposite

16. zepdrix

yepppppp

17. mattlai7

5/-2sqrt6

18. zepdrix

yay good job! \c:/

19. zepdrix

For your final answer it might be a good idea to rationalize your answer. Get the irrational number out of the denominator. Multiply the top and bottom by sqrt6.

20. mattlai7

5sqrt6/-12?

21. zepdrix

Yes.

22. mattlai7

a little confused on the denominator part....not sure if its 12 or not

23. zepdrix

$\large 2 \cdot \sqrt6 \sqrt6 \qquad = \qquad 2\cdot6 \qquad = \qquad 12$Hmm yah everything looks ok there!

24. mattlai7

ok thanks!!!

25. mattlai7

thanks for the help some time a 8am class only aborbs so much in an hour and with 300 classmates lol

26. zepdrix

heh ill bet! XD