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Trig function help

  1. Jan 25, 2015 #1

    Eric Manuel

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    thanks
     
    Last edited: Jan 25, 2015
    Eric Manuel, Jan 25, 2015
  2. jcsd
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  3. Jan 25, 2015 #2

    Jimmy Johnson

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    Can you give the entire problem?
     
    Jimmy Johnson, Jan 25, 2015
  4. Jan 25, 2015 #3

    Eric Manuel

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    all thats given is that
     
    Eric Manuel, Jan 25, 2015
  5. Jan 25, 2015 #4

    MostlyHarmless

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    What is the question? What are you asked to find?
     
    MostlyHarmless, Jan 25, 2015
  6. Jan 25, 2015 #5

    Eric Manuel

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    what is =cos2x
     
    Eric Manuel, Jan 25, 2015
  7. Jan 25, 2015 #6

    MostlyHarmless

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    What is x supposed to be? Is the given problem in terms of t? Are you trying to solve for t and then plug that value in for x?
     
    MostlyHarmless, Jan 25, 2015
  8. Jan 25, 2015 #7

    Eric Manuel

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    oh my bad its cos2t

    its basically asking you to use the double angle formula
    but im not sure how to find those values
     
    Eric Manuel, Jan 25, 2015
  9. Jan 25, 2015 #8

    Jimmy Johnson

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    If you're simply trying to find t, then use cot = 1/tan... If you have a value for t why do you need to use the double angle identities?
     
    Jimmy Johnson, Jan 25, 2015
  10. Jan 25, 2015 #9

    Eric Manuel

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    to find what cos2t is ....

    you can find what tan is from cot
    then i drew a triangle and got the hypot to be sqrt(41)

    but how do i do that math

    formula for cos2t = cos^2(t) - sin^2(t)
     
    Eric Manuel, Jan 25, 2015
  11. Jan 25, 2015 #10

    Jimmy Johnson

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    Are you forbidden from using a calculator?
     
    Jimmy Johnson, Jan 25, 2015
  12. Jan 25, 2015 #11

    Eric Manuel

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    teacher said u can use it
    but need to know how to do it without one for test
     
    Eric Manuel, Jan 25, 2015
  13. Jan 25, 2015 #12

    Jimmy Johnson

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    Well that may have been useful information at the start...

    you have cot = -5/4, find what tan is then recall tan = sin/cos ... or recall cot = cos/sin...
     
    Jimmy Johnson, Jan 25, 2015
  14. Jan 25, 2015 #13

    Eric Manuel

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    well i worked it out and i believe i got the answer to be 9/41
     
    Eric Manuel, Jan 25, 2015
  15. Jan 25, 2015 #14

    Jimmy Johnson

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    That's what I'm getting
     
    Jimmy Johnson, Jan 25, 2015
  16. Jan 25, 2015 #15

    Jimmy Johnson

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    Jimmy Johnson, Jan 25, 2015
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