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CosX in terms of Tanx?

  1. Jan 30, 2007 #1
    1. The problem statement, all variables and given/known data
    seems simple, but i am stumped. Says write cos(x) in terms of tan(x).


    2. Relevant equations
    would this be a reciprocal equation? or a Pythagorean? I'm lost


    3. The attempt at a solution

    i dont even know where to begin.
     
  2. jcsd
  3. Jan 30, 2007 #2

    mathwonk

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    sin/cos = tan, so cos= sin/tan. har har.

    can you use derivatives?
     
  4. Jan 30, 2007 #3
    Write down the two formula for tan x and cos x for a right angle triangle. Are there any similar terms in those equations?

    Edit: Beaten to it.
     
  5. Jan 30, 2007 #4

    mathwonk

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    DO YOU KNOW WHaT TAN' IS? or 1 + tan^2?
     
  6. Jan 30, 2007 #5
    the angle is unknown. I think thats why its confusing me.

    sin/cos = tan, so cos= sin/tan - these are what i have. But would that be the answer? tan= sin/cos ? or cos=sin/tan?
     
  7. Jan 30, 2007 #6

    mathwonk

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    i was joking. read my second post.
     
  8. Jan 30, 2007 #7
    :smile: now im even more confused.

    would it be cos=sin/tan?
     
  9. Jan 30, 2007 #8

    arildno

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    How can you express sine in terms of cosine?
     
  10. Jan 30, 2007 #9
    i dont know :confused:
     
  11. Jan 30, 2007 #10

    arildno

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    Well, what RELATION exists between the sine and cosine of an angle?
     
  12. Jan 30, 2007 #11

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    [itex]\sin x= \sqrt{1-cos^2x}[/itex]


    :devil:
     
    Last edited: Jan 30, 2007
  13. Jan 30, 2007 #12
    hmm, sin/cos=tan, cos/sin=cot, sin^2 + cos^2=1

    i need cos(theta) in terms of tan(theta) though. Unless thats what we are working up to :)
     
  14. Jan 30, 2007 #13

    arildno

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    So, look at the last identity you posted.

    What do you get by dividing ôn both sides with cos^{2} ?
     
  15. Jan 30, 2007 #14
    sin^2 = 1/cos^2?
     
  16. Jan 30, 2007 #15

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  17. Jan 30, 2007 #16

    arildno

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    Don't you know how to divide an equation with a number?
     
  18. Jan 30, 2007 #17
    you can express sine in terms of cosine as cos (x-90) where x is in degrees or in radians cos (x-pi/2).
     
  19. Jan 30, 2007 #18
    Am I the first person who thinks it can't be done? Maybe I'm overlooking something, but I'm seeing a sign problem. (+/- when you solve)
     
  20. Jan 30, 2007 #19
    Only works for 1st and 2nd quadrant angles, that is, angles between 0 and 180 degrees. (or between 0 and 2Pi). Plus, it works for 0 degrees and 180 degrees. If you're in the 3rd or 4th quadrant, then you'd have to use a negative square root.
     
    Last edited: Jan 30, 2007
  21. Jan 30, 2007 #20

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    yes,
    how to use +- in latex?
     
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