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Limit of Trig Function

  1. Oct 18, 2005 #1
    Hello everyone,

    I am having some trouble finding the
    limit as x approaches pi/4 of (sinx-cosx)/(cos2x)

    I can't really seem to get started on this one. I'm horribe at this sort of thing. Do you think it would be beneficial to break up the cos2x into one of its formulas or would it just be a waste of time? I've tried leaving it as cos2x and I've tried it with its identities but nothing is working for me. I can't get anything to simplify. Any help would be greatly appreciated as I have no idea where to go with this. Thanks.
     
  2. jcsd
  3. Oct 18, 2005 #2

    Tom Mattson

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    Your goal here is to learn mathematics, not get the answer as quickly as possible. So if you try playing around with trig identities and you learn that an approach doesn't work, then how could that possibly be considered a waste of time? That's how you get good at this sort of thing.

    Or maybe just less horrible at it. :biggrin:

    Try something, and if you get stuck present what you've tried and we will help you. Them's the rules.
     
  4. Oct 18, 2005 #3
    I'm sorry I think my post came off a bit wrong, I didn't mean it to sound like I just wanted a quick answer. I had actually tried subsituting the identites in and seeing if I could get any further with them before I made the actual post. My problem is that once I subsitute the identies in I don't know where to go with it. The identities I tried subsituting were the ones for cos2x, so I put in both 2cos^2 (x)-1 and 1-2sin^2(x). And I won't lie, that's are far as I got, I just don't see where to take them from there. I don't need you to lay the answer out for me, I think I just need a bit of a boost to get going. Sorry for any misunderstanding.
     
  5. Oct 19, 2005 #4

    TD

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    It's a good idea to substitute cos(2x), but the two identities you used are both derived from another, 'third' (so actually 'first') identity, i.e. [itex]\cos \left( {2x} \right) = \cos ^2 x - \sin ^2 x[/itex]

    Do you see what you could do next?
     
  6. Oct 19, 2005 #5
    Oh I actually wasn't aware that was an identity for cos2x. I think I can see something to do from here, I will just go try it and see how it works, if I can't get anywhere i'll post back with my steps. Thanks a lot.
     
  7. Oct 19, 2005 #6
    Ok, so what I did was I put the identity for cos2x in the denominator and factored it to (cosx+sinx)(cosx-sinx). Then I broke up the limit into two parts, one part was (sinx)/(cosx-sinx) and the other was (cosx)/(cosx+sinx). When I solved for by substituing x=pi/4, I ended up with 0-2, making the limit equal -2. I checked this on my calculator and it did not work, unless I typed it in wrong but I don't think I did. What could I have done wrong?
     
  8. Oct 19, 2005 #7
    [itex]\cos \left( {2x} \right) = \cos ^2 x - \sin ^2 x = (cos x - sin x)(cos x + sin x)[/itex]
    does it ring a bell now? You have to do something with the numerator.
     
  9. Oct 19, 2005 #8

    TD

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    You already factored correctly, but splitting is not such a good idea now. Follow iNCREDiBLE's advice and take a look at the entire fraction now!

    PS: that identity for cos(2x) can be easily found by using the som formula for cos(x+y) with y = x :smile:
     
  10. Oct 19, 2005 #9

    VietDao29

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    Yup. This is correct.
    Nope. This is wrong. You should note that:
    [tex]\frac{(\sin x - \cos x)}{(\sin x + \cos x)(\cos x - \sin x)} \neq \frac{\sin x}{(\cos x - \sin x)} - \frac{\cos x}{(\sin x + \cos x)}[/tex]
    You have sin x - cos x in the numerator and - (sin x - cos x) (sin x + cos x) in the numerator. What should you do to get rid of sin x - cos x?
    Viet Dao,
     
  11. Oct 19, 2005 #10
    Oh I feel like an absolute idiot, the numerator and one of the factors in the denominator cancel, duh! Sorry about that guys, major blonde moment.
     
  12. Oct 19, 2005 #11

    TD

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    No problem, you got it now I assume? :smile:
     
  13. Oct 19, 2005 #12
    Ok so I canceled out the common factors and was left with -1/(cosx+sinx) which when the value for x was substituted leaves you with -1/(2/square root2) which when simplified gives a limit of -squareroot2/2?
     
  14. Oct 19, 2005 #13

    TD

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    Looks OK, apart from the minus in the nominator, where did that come from?
     
  15. Oct 19, 2005 #14
    I factored out a negative to make it (-sinx+cosx)- which equals -(cosx-sinx) so that it could be factored out. Is this wrong to do?
     
  16. Oct 19, 2005 #15

    TD

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    Oh yes, totally correct :smile:

    I had cos(x)-sin(x) in the numerator, my bad!
     
  17. Oct 19, 2005 #16
    Oh ok, it is right then!!! Thanks a lot, I really appreciate your help and putting up with my retarded mistakes.....haha!
     
  18. Oct 19, 2005 #17
    scorpa, don't be so harsh on yourself. :wink:
     
  19. Oct 19, 2005 #18
    Lol, I just don't like making mistakes, especially stupid ones.
     
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