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Homework Help: If tan x = k(where k is a positive constant),How is tan( pie/2 − x)=1/k

  1. May 6, 2012 #1
    1. The problem statement, all variables and given/known data
    The exact question is in question one part ii of

    http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_qp_11.pdf

    2. Relevant equations

    3. The attempt at a solution
    For the first one I just pluged in a value for x, and got -k. But I got stuck a two which tells me my approach is perhaps not entirely what cie is looking for, is there a better way of getting the answer?(like drawing quadrants or something)
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. May 6, 2012 #2
    The standard way would be to simply write the tangent function as the ratio of sine and cosine, and then use the relationship between those two to rewrite, say, [itex]\text{sin}(\frac{\pi}{2}-x)[/itex] in terms of [itex]\text{cos}(x)[/itex].
  4. May 6, 2012 #3
    I dont quite get you, although I know tanx=sinx/cosx I am not sure how I express ∏/2 -x, in terms of sin and cos. Do you mean, I should write sinx/tanx= K and then proceed? But wont that just introduce more variables?
  5. May 6, 2012 #4
    I mean you should write [tex]\text{tan}(\frac{\pi}{2}-x) = \frac{\text{sin}(\frac{\pi}{2}-x)}{\text{cos}(\frac{\pi}{2}-x)}[/tex] and continue from there, simplifying the numerator and denominator.
  6. May 6, 2012 #5
    Im terribly sorry Mr.Dan, but I have avery thick head, and im trying, but I dont seem to understand, I did as you said, I got sin(∏-2x)/2 divided by cos(∏-2x). But that just leads me back to sin(∏-2x)/cos(∏-2x) !
  7. May 6, 2012 #6
    It's actually simpler than that. I'm not talking about, say, the double angle formula, I'm talking about how, for example [tex]\text{sin}(\frac{\pi}{2}-x) = \text{cos}(x).[/tex]
  8. May 6, 2012 #7
  9. May 6, 2012 #8
    Alright, okay. Thanks a lot, I thnk Im gettin it. Thanks a tonne
  10. May 6, 2012 #9
  11. May 6, 2012 #10


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    Homework Helper


    [tex]\sin\left(A+B\right)\neq \frac{\sin\left(2A+2B\right)}{2}[/tex]
  12. May 7, 2012 #11


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    Staff: Mentor

    Hi mutineer123! Draw a right-angled triangle, and mark one of the acute angles x. Label the lengths of the two sides appropriate to this question such that the ratio of these sides = k.

    The remaining angle in the triangle is (π/2 -x) ....
  13. May 7, 2012 #12
    Holy CRAP! I did not think of a right angled triangle!!! It made it so much easier! THANKS!
  14. May 7, 2012 #13
    we havent learnt that yet.
  15. May 8, 2012 #14


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    Homework Helper

    Ok, but that looks to be essentially what you've done earlier:
    You either meant [tex]\sin\left(\frac{\pi}{2}-x\right)=\frac{\sin\left(\pi-2x\right)}{2}[/tex] or [tex]\frac{\sin\left(\frac{\pi}{2}-x\right)}{\cos\left(\frac{\pi}{2}-x\right)}=\frac{\sin\left(\frac{\pi-2x}{2}\right)}{\cos\left(\frac{\pi-2x}{2}\right)}=\frac{\sin\left(\pi-2x\right)}{\cos\left(\pi-2x\right)}[/tex]

    Which are both incorrect!
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