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

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1. zepdrix Group Title

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

2. mattlai7 Group Title

not quite sure why?

3. zepdrix Group Title

|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 Group Title

ok now i understand that part

5. zepdrix Group Title

|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 Group Title

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

7. mattlai7 Group Title

|dw:1362773386830:dw|

8. zepdrix Group Title

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 Group Title

opposite will equal 4?

10. mattlai7 Group Title

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

11. zepdrix Group Title

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 Group Title

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

13. zepdrix Group Title

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 Group Title

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

15. mattlai7 Group Title

hypotenuse over opposite

16. zepdrix Group Title

yepppppp

17. mattlai7 Group Title

5/-2sqrt6

18. zepdrix Group Title

yay good job! \c:/

19. zepdrix Group Title

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 Group Title

5sqrt6/-12?

21. zepdrix Group Title

Yes.

22. mattlai7 Group Title

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

23. zepdrix Group Title

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

24. mattlai7 Group Title

ok thanks!!!

25. mattlai7 Group Title

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

26. zepdrix Group Title

heh ill bet! XD