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Prove the following identity algebraically

  1. Mar 22, 2009 #1
    1. The problem statement, all variables and given/known data
    PhysicsForumcommarch21sttrig.png

    Must be proven algebraically, duh!
    2. Relevant equations
    trig identities


    3. The attempt at a solution

    PhysicsForumcommarch21sttrigpart2.png

    PhysicsForumcommarch21sttrigpart3.png

    I'm at a loss as what to do next. Any help would be appreciated.
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Mar 22, 2009 #2

    rock.freak667

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    try using [itex]sin^2x+cos^2x=1[/itex] and substitute for cos2x
     
  4. Mar 22, 2009 #3

    symbolipoint

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    Another approach, from the lefthand side of the original given equation, do you see that
    (csc2x-cot2x)2=the expression originally on the leftside.
     
    Last edited: Mar 22, 2009
  5. Mar 22, 2009 #4

    symbolipoint

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    Yes, again, do as in #3. Then do the substitutions for the meaning of csc and for cot, simplify using algebra steps, and use the identity suggested in post #2. This becomes very uncomplicated.
     
  6. Mar 22, 2009 #5
    [tex] \frac{2 - cos^2x + 2cos^4x}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

    How am I going to get rid of the two on the L.H.S.? I can't just directly cancel it, because of the [tex] - cos^2x [/tex] right?

    I'm at a loss, even with all this help. So I'm going to show you my next step and maybe you people can tell me what I'm doing wrong.

    [tex] \frac{2 - [cos^2x ( 1 + 2cos^2x )]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

    [tex] \frac{2 - [cos^2x ( 1 + 2(1 - sin^2x)]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

    [tex] \frac{2 + cos^2x - (2sin^2xcos^2x)}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

    I do this next: I cancel the two coefficients in the denominator and in the bracketed term, then I subtract the exponent from the [tex]sin^2x[/tex] in the brackets from the denominator.

    I'm left with this:

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

    which simplifies to:

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

    Where did I go wrong?
     
  7. Mar 22, 2009 #6
    Let me restart from here:

    [tex] \frac{2 - [cos^2x ( 1 + 2(1 - sin^2x)]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

    [tex] \frac{2 - [cos^2x ( 1 + 2 - 2sin^2x)]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

    [tex] \frac{2 - [3cos^2x - 2sin^2xcos^2x]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

    [tex]\frac{2 - 3cos^2x + 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

    In the previous step I subtracted the exponents for sin in the denominator from the numerator. Is this a legal maneuver?

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

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

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

    Is this correct? Thanks again for the help guys.
     
    Last edited: Mar 22, 2009
  8. Mar 22, 2009 #7

    I made in error in writing down my work, it should read:

    [tex]\frac{2 - 3cos^2x + 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

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

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

    [tex]\frac{1-cos^2x}{sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]
     
  9. Mar 22, 2009 #8
    I would like to point out an error you made very early on.

    You started with:
    [tex]csc^{4}x - 2*csc^{2}x*cot^{2}x + cot^{4}x = 1[/tex]

    And somehow got to:
    [tex]\frac{1}{sin^{4}x} - \frac{cos^{2}x}{2*sin^4{x}} + \frac{cos^{4}x}{sin^{4}x} = 1[/tex]

    How did your coefficient of the second term flip from the numerator to the denominator? It should read:
    [tex]\frac{1}{sin^{4}x} - \frac{2*cos^{2}x}{sin^4{x}} + \frac{cos^{4}x}{sin^{4}x} = 1[/tex]

    From there it is pretty straightforward. Add the terms together, factor the top, then use the Pythagorean trig identities to reduce it to 1.
     
  10. Mar 22, 2009 #9
    Thanks for the constructive criticism.
     
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