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Homework Help: Trig identity

  1. Mar 6, 2006 #1
    how can i prove

    tan X/sinx+cosx=sin^2 X + sinXcosX/cos X - 2cos^3X

    so far i've tried using basic identities to figure it out, but i just end up getting confused.


    :confused:
     
    Last edited: Mar 6, 2006
  2. jcsd
  3. Mar 6, 2006 #2
    The way you have written it is confusing. Are there any paranthesis missing?
     
  4. Mar 6, 2006 #3
    no..

    hmm lemme write it out another way

    how can i prove

    tanX
    ______
    sinx+cosx

    =

    sin^2 X + sinXcosX
    ________________
    cos X - 2cos^3X
     
  5. Mar 6, 2006 #4

    CarlB

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    Try putting x = pi/4 and see what you get.

    Carl
     
  6. Mar 6, 2006 #5
    In other words, there were parenthesis missing...
     
  7. Mar 7, 2006 #6

    VietDao29

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    The problem is wrong. Are you sure you copied this correctly?
    I think it should be:
    [tex]\frac{\tan x}{\sin x + \cos x} = \frac{\sin ^ 2 x + \sin x \cos x}{\cos x + 2 \sin x \cos ^ 2 x}[/tex]
     
  8. Mar 7, 2006 #7
    i copied it as it was exactly written in the book.. :confused:
     
  9. Mar 7, 2006 #8

    arildno

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    Dearly Missed

    No, you haven't copied this correctly: Put X=0
    Left hand side yields 2, whereas right-hand side yields -2.

    Read your book again; this time with your eyes open.
     
  10. Mar 7, 2006 #9
    Try cancelling the common factors and try compreesing the question to its simplest form. You would then probably then get a breakthrough
     
  11. Mar 7, 2006 #10
    i am positive i copied it correctly. i would take a picture from the book, but i don't have a camera or scanner right now.
     
  12. Mar 7, 2006 #11

    Hootenanny

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    When you put x = 0 in both sides are equal to zero.

    Just rewritting what clook posted in a more organised manor:

    [tex]\frac{\tan x}{\sin x + \cos x} = \frac{\sin^2 x + \sin x \cos x}{\cos x - 2\cos^3 x}[/tex]
     
  13. Mar 7, 2006 #12
    This statement is equivalent to saying [tex]sinx=-cosx[/tex], so something is VERY fishy here.

    -Dan
     
  14. Mar 7, 2006 #13
    Ok clook, assuming that

    [tex]\frac{\tan x}{\sin x + \cos x} = \frac{\sin^2 x + \sin x \cos x}{\cos x - 2\cos^3 x}[/tex]

    is what you have in your book. I worked with the RHS and got this:

    [tex]\frac{\tan x}{\sin x-\cos x}[/tex]

    As you can see, this is different from the LHS in your equation. Something is indeed fishy here.
     
  15. Mar 8, 2006 #14

    VietDao29

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    Okay, look at your problem again, does it say something like:
    Prove this identity, or does it tell you to solve the equation?
     
  16. Mar 8, 2006 #15

    Hootenanny

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    Yes solve would make alot more sense (and would be very easy as well) :wink:
     
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