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Homework Help: Trig identity with natural logs and absolute value?

  1. Jan 12, 2012 #1
    Trig identity with natural logs and absolute value??

    1. The problem statement, all variables and given/known data
    -ln|csc(x) + cot(x)|= ln|cscx(x)-cot(x)|


    2. Relevant equations



    3. The attempt at a solution
    I got that csc(x)=1/sin(x) and cot(x)=cos(x)/sin(x), giving me a common denominator, added together I have 1+cos(x)/sin(x). So now the Absolute value and ln are throwing me a curve, and I'm stuck at -ln|1+cos(x)/sin(x)| = ln|1-cos(x)/sin(x). Even if you can just point me in the right direction it would be a big help, thanks in advance!!
     
  2. jcsd
  3. Jan 12, 2012 #2
    Re: Trig identity with natural logs and absolute value??

    Try to get rid of the ln.
     
  4. Jan 12, 2012 #3
    Re: Trig identity with natural logs and absolute value??

    You should be able to get rid of the absolute value signs partly by just looking at intervals 0<=x<=1/2 pi etc. But it seems like there will be trouble since the argument of the logarithm vanishes at some points probably worth a note. you have -ln at one side maybe you should taking the inverse of the argument there.
     
  5. Jan 12, 2012 #4
    Re: Trig identity with natural logs and absolute value??

    Haha right or just do that
     
  6. Jan 12, 2012 #5
    Re: Trig identity with natural logs and absolute value??

    how do you get rid of the ln? does it have to do with multiplying by the inverse?
     
  7. Jan 12, 2012 #6
    Re: Trig identity with natural logs and absolute value??

    Also remember the rules for logarithms.

    [itex]lnx^{k}=klnx[/itex]
     
  8. Jan 12, 2012 #7
    Re: Trig identity with natural logs and absolute value??

    not sure how to relate lnx^k to this problem.. is x the function? what is k?
     
  9. Jan 12, 2012 #8
    Re: Trig identity with natural logs and absolute value??

    I meant it as a general rule for logarithms. In this problem I mean.

    [itex]-ln|cosecx + cotx| = ln(cosecx + cotx)^{-1}[/itex]
     
  10. Jan 12, 2012 #9

    Curious3141

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    Re: Trig identity with natural logs and absolute value??

    Hints:

    [itex]\ln x^k = k\ln x[/itex]. What happens when [itex]k = -1[/itex]?

    [itex]\frac{a}{b+c} = \frac{(a)(b-c)}{(b+c)(b-c)}[/itex]. This form will come in useful. Simplify the denominator and see how you can apply a trig identity to make it really simple.
     
  11. Jan 12, 2012 #10
    Re: Trig identity with natural logs and absolute value??

    when k = -1 does that mean take the inverse, like ln|1+cos(x)/sin(x)|^-1 ?
     
  12. Jan 12, 2012 #11

    Curious3141

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    Re: Trig identity with natural logs and absolute value??

    Yes. Remember it's the logarithm of the inverse, NOT the inverse of the whole logarithm.

    It's clearer if you express the inverse as 1/(something). Leave everything in terms of cot and csc, and apply the second hint I gave you.
     
  13. Jan 12, 2012 #12
    Re: Trig identity with natural logs and absolute value??

    [itex]ln(cosecx + cotx)^{-1} = ln|\frac{1}{cosecx + cotx}|[/itex]

    So now you know that

    [itex]ln|\frac{1}{cosecx + cotx}| = ln|cosecx - cotx|[/itex]
     
  14. Jan 12, 2012 #13

    Curious3141

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    Re: Trig identity with natural logs and absolute value??

    Bread, it's not an equation he's supposed be solving. He's supposed to prove a trig identity. So he's supposed to manipulate one side (in this case, starting from the LHS is fine) till it equals the other. He's not supposed to start by assuming equality.
     
  15. Jan 12, 2012 #14
    Re: Trig identity with natural logs and absolute value??

    Ah k, I wasn't too sure what he was trying to do, he didn't really state what the question said.

    Though I do now have a question about that, if I have a question that says show a=b, can't i do so by showing that they both = c?
     
  16. Jan 12, 2012 #15
    Re: Trig identity with natural logs and absolute value??

    OK! got it now, the -ln makes me take the inverse by raising the power to negative 1, then I put 1/csc(x)+cot(x) then the 1 on top goes away when I switch the sign on the bottom to negative. Awesome! So i didn't need to worry about the absolute value, or figuring out that csc(x)+cot(x) equaled 1 + cos(x)/sin(x). Thanks again!!!
     
  17. Jan 12, 2012 #16

    Curious3141

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    Re: Trig identity with natural logs and absolute value??

    It's in the thread title.

    You can do that, but it's usually preferable to try to manipulate one side until it becomes the other.

    Sometimes, to "see" the solution, you use the method you alluded to (to simplify both to the same form), but once you do this, you can reverse engineer one side of it so that you can work from LHS to RHS or vice versa when you present the final proof.
     
  18. Jan 12, 2012 #17

    Curious3141

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    Re: Trig identity with natural logs and absolute value??

    Not so fast. :smile: You'd better show exactly what you did, because something about what you wrote doesn't seem right.
     
  19. Jan 12, 2012 #18
    Re: Trig identity with natural logs and absolute value??

    What exactly did you do?
     
  20. Jan 12, 2012 #19
    Re: Trig identity with natural logs and absolute value??

    I generally work until they reach a common solution, like this I showed they were both equal to [itex]sin^{2}x[/itex]
     
  21. Jan 12, 2012 #20

    Curious3141

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    Re: Trig identity with natural logs and absolute value??

    Actually, there's a much more direct way here. Can you work it out with the hints I provided?

    BTW, I don't think either side is equal to [itex]\sin^2{x}[/itex].
     
  22. Jan 12, 2012 #21
    Re: Trig identity with natural logs and absolute value??

    I have done it another way, I think it's what he is saying.

    [itex]\frac{1}{cosecx + cotx}\times\frac{cosecx - cotx}{cosecx - cotx}[/itex]

    [itex]= \frac{cosecx - cotx}{\frac{sin^{2}x}{sin^{2}x}}[/itex]

    [itex]=cosecx - cotx[/itex]
     
  23. Jan 12, 2012 #22

    Curious3141

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    Re: Trig identity with natural logs and absolute value??

    Yup, that's right. For the denominator, it's acceptable to recognise the trig identity [itex]\csc^2{x} - \cot^2{x} = 1[/itex] directly. But it would've been good if the OP had worked this out for himself. Right now, neither of us is sure what he did.
     
  24. Jan 12, 2012 #23
    Re: Trig identity with natural logs and absolute value??

    ok guys, sorry, the inverse of csc(x)+cot(x) is 1/csc(x)+cot(x), and I saw the 1 on top and just thought it meant switch csc(x)+cot(x) to csc(x)-cot(x), making it equal to the other side, proving the identity, isn't that right?
     
  25. Jan 12, 2012 #24

    Curious3141

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    Re: Trig identity with natural logs and absolute value??

    Good thing you posted.

    A couple of things:

    1) Brackets are important. [itex]\frac{1}{a} + b \neq \frac{1}{a+b}[/itex]. When you write "1/csc(x)+cot(x)", it's generally read as the former, while you meant the latter. So please be careful about placing your brackets to render the right expression.

    2) In general, [itex]\frac{1}{a+b} \neq a - b[/itex]!!! It's true in this case because of the particular nature of those trig ratios, but you have to *show* that it's true in this particular case. But to see why it isn't true in general, try a = 3, b = 2.
     
  26. Jan 12, 2012 #25
    Re: Trig identity with natural logs and absolute value??

    ok, yeah I need to learn how you guys post on here in math terms instead of the keyboard, sorry for the confusion of 1/csc(x)+cot(x) being read as (1/csc(x))(cot(x)).. and yeah I just got lucky, because I didn't even think to multiply the top and bottom by the denominator. Thanks again!! P.S. is it pushing it to post twice in 1 night? I got 1 more problem I need answered but I feel like that's frowned on around here, thanks again!!!!
     
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