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Homework Help: Trigonometric Proof Help

  1. Jan 13, 2009 #1
    1. The problem statement, all variables and given/known data

    (tan2x) - (sin2x) = (tan2x)(sin2x)

    Prove these two are equal.

    2. Relevant equations

    Various trig identities, mainly tan2x = sin2x / cos2x

    3. The attempt at a solution

    I tried putting in Sin2x / cos2x for tan, but I don't know where to go from there.

    Anyone mind helping me out?
     
  2. jcsd
  3. Jan 13, 2009 #2

    jgens

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    This has a really simple solution. Factor out sin(x)^2 from the left hand side and then simplify.
     
  4. Jan 13, 2009 #3
    I don't how/where you can factor sin(x)^2 out in this case. This is what I have done to the left side thus far.

    [tex]\frac{sin^{2}x}{cos^{2}x} - \frac{sin^{2}x}{1}[/tex]

    then

    [tex]\frac{sin^{2}x}{cos^{2}x} - \frac{sin^{2}xcos^{2}x}{cos^{2}x}[/tex]

    which ends up as

    [tex]\frac{sin^{2}x - sin^{2}xcos^{2}x}{cos^{2}x}}[/tex]
     
  5. Jan 13, 2009 #4

    jgens

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    Well, using your steps, you're almost at the solution. Again, factor sin(x)^2 out of the numerator and simplify. :)

    This is what I was talking about in my initial post: tan(x)^2 - sin(x)^2 = sin(x)^2 (sec(x)^2 - 1), which is easy to simplify.
     
  6. Jan 13, 2009 #5
    [tex]\frac{sin^{2}x (1 -cos^{2}x)}{cos^{2}x}}[/tex]

    I'm missing something, because I still don't get it. :grumpy: What am I overlooking? Sin/Cos is equal to Tan, but then I have (1 - cos2x)/(cos2x)
     
  7. Jan 13, 2009 #6

    jgens

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    Are you familiar with the trig. identity sin(x)^2 + cos(x)^2 = 1? If you are, the solution should seem fairly obvious in both instances.
     
  8. Jan 13, 2009 #7
    At this point I have:

    [tex]\frac{sin^{2}x (1 -cos^{2}x)}{cos^{2}x}} = (\frac{sin^{2}x}{cos^{2}x})(\frac{sin^{2}x}{1}}) [/tex]

    I know that identity, but I can't see how it can be used on either side.

    I'm sorry for my complete lack of ability to grasp this.
     
  9. Jan 13, 2009 #8

    jgens

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    If you know that identity then you surely know 1 - cos(x)^2 = sin(x)^2 or similarly that sec(x)^2 - 1 = tan(x)^2.
     
  10. Jan 13, 2009 #9
    Just realized this:

    [tex]\frac{sin^{2}x (1 -cos^{2}x)}{cos^{2}x}} = \frac{sin^{2}x(sin^{2}x)}{cos^{2}x}[/tex]

    which turns into


    [tex]\frac{sin^{2}x (1 -cos^{2}x)}{cos^{2}x}} = \frac{sin^{2}x(1-cos^{2}x)}{cos^{2}x}[/tex]

    Did I do anything incorrect there? And thank you very much for your help.
     
  11. Jan 13, 2009 #10

    jgens

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    Nothing there is incorrect - in fact that result should tell you something - but you should have made a connection which it appears you have still failed to make. If 1 - cos(x)^2 = sin(x)^2 and we have (1 - cos(x)^2)/cos(x)^2, what substitution can I make to simplify the expression?
     
  12. Jan 13, 2009 #11
    sin(x)^2
    ------------ = tan(x)^2
    cos(x)^2

    Correct? I get confused when this happens though:

    [tex]\frac{sin^{2}x (sin^{2}x)}{cos^{2}x}}[/tex]

    to

    [tex]tan^{2}x\frac{sin^{2}x}{cos^{2}x}}[/tex]

    or does the cos(x)^2 drop?
     
  13. Jan 13, 2009 #12

    jgens

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    The cos(x)^2 term does not drop but is merely factored out essentially in the tan(x)^2 term; hence, your resulting equation should read tan(x)^2 sin(x)^2. Q.E.D.
     
  14. Jan 13, 2009 #13
    Okay, I understand. Thank you a lot for the great help. :)
     
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