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mattlai7

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

    • one year ago
  2. mattlai7
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    not quite sure why?

    • one year ago
  3. zepdrix
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    |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.

    • one year ago
  4. mattlai7
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    ok now i understand that part

    • one year ago
  5. zepdrix
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    |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 \(\large t\).

    • one year ago
  6. zepdrix
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    \(\large \cos t=-\dfrac{1}{5}=\dfrac{adjacent}{hypotenuse}\) So how would we label these sides? HINT: We always let the hypotenuse be positive.

    • one year ago
  7. mattlai7
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    |dw:1362773386830:dw|

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

    • one year ago
  9. mattlai7
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    opposite will equal 4?

    • one year ago
  10. mattlai7
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    sorry did an error......missing side equals to \[+/- 2\sqrt{6}\]

    • one year ago
  11. zepdrix
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    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?

    • one year ago
  12. mattlai7
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    we want a negative answer because tan is only positive in the 3rd Q

    • one year ago
  13. zepdrix
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    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|

    • one year ago
  14. zepdrix
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    Ok good. Everything is labeled correctly. Just have to find \(\large \csc t\) now! :)

    • one year ago
  15. mattlai7
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    hypotenuse over opposite

    • one year ago
  16. zepdrix
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    yepppppp

    • one year ago
  17. mattlai7
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    5/-2sqrt6

    • one year ago
  18. zepdrix
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    yay good job! \c:/

    • one year ago
  19. zepdrix
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    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.

    • one year ago
  20. mattlai7
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    5sqrt6/-12?

    • one year ago
  21. zepdrix
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    Yes.

    • one year ago
  22. mattlai7
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    a little confused on the denominator part....not sure if its 12 or not

    • one year ago
  23. zepdrix
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    \[\large 2 \cdot \sqrt6 \sqrt6 \qquad = \qquad 2\cdot6 \qquad = \qquad 12\]Hmm yah everything looks ok there!

    • one year ago
  24. mattlai7
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    ok thanks!!!

    • one year ago
  25. mattlai7
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    thanks for the help some time a 8am class only aborbs so much in an hour and with 300 classmates lol

    • one year ago
  26. zepdrix
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    heh ill bet! XD

    • one year ago
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