1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Mentallic

    User Avatar
    Homework Helper

    Noo...

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

    NascentOxygen

    User Avatar

    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

    Mentallic

    User Avatar
    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: If tan x = k(where k is a positive constant),How is tan( pie/2 − x)=1/k
Loading...