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Trig question

  1. Jan 2, 2008 #1
    [SOLVED] Trig question

    Hi, I'm working on some trig questions and can't get the answer for this last question; I have a hunch that it might be one of those obvious ones where you think too much :uhh: ....

    Either way, help would be much appreciated!

  2. jcsd
  3. Jan 2, 2008 #2


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    Homework Helper

    PF welcome.
    ok, you should really tell us what you have tried.
    hint: if [tex]\sin \,x \neq 0[/tex] you can divide through and eliminate the cos and sin... and put them into another trig function.
    when sin x =0 then it is simple eh?
    (same applies to cos)
  4. Jan 2, 2008 #3
    Thanks mjsd,

    I've established that it can be simplified to cos2x-sin2x=0, but I'm not sure where to go from there... unless cos2x and sin2x are both zero?

    How can you divide through to eliminate cos and sin?
  5. Jan 2, 2008 #4
    You might also consider temporarily substituting a for cosx and b for sinx, and see if you get anything that looks easier to solve.
  6. Jan 2, 2008 #5
    What about trying mjsd's hint in the last equation: divide by [tex]\sin2\,x[/tex] if [tex]\sin2\,x \neq 0[/tex]


    change [tex] \cos2\,x[/tex] to [tex]\sin(\frac{\pi}{2}-2\,x)[/tex].
  7. Jan 2, 2008 #6


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    Staff Emeritus
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    To continue belliot4488's suggestion, can you solve [itex]a^2- 2ab- b^2= 0[/itex] for a in terms of b?
  8. Jan 2, 2008 #7
    oh! I think I've got it:

    divide both sides by cos2x to get tan2x=1
    work out four answers to be π/8, 5π/8, 9π/8, and 13π/8

    is this correct?

    Thanks for everyone's quick replies! :D
  9. Jan 2, 2008 #8
    Correct! :smile:
  10. Jan 2, 2008 #9
    Thank you so much!
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