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Math Help - Proving Another Identity

  1. #1
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    Proving Another Identity

    I'm stuck again... I don't know why I am having so much trouble with these. There is clearly something I'm just not seeing.

    1 + cos2x = cotx sin2x

    I think I need to simplify the right hand side?

    This is what I've done, but I'm making a real mess of things....

    cotx sin2x
    = (cosx/sinx)(sin2x)
    = cosx sin2x/ sinx
    = [cosx (2sinx cosx)] / sinx

    And then I'm lost...
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  2. #2
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    In problems such as this, the first thing I would do is get all trig functions to have the same argument. I never know what I have until I do that. Use the double-angle formulas to get rid of the 2x arguments. What do you get?
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  3. #3
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    On the left hand side I have:

    1 + cos2x
    = 1 + 2cos^2 - 1
    = 2cos^2x


    On the right hand side I have:

    cotx sin2x
    = cosx/sinx (sin2x)
    = (cosx/sinx) (2sinx cosx / 1)
    = 2sinx 2cos^2x / sin x
    = sinx 2cos^2x

    But that's clearly not right....
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  4. #4
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    Quote Originally Posted by starshine84 View Post
    On the left hand side I have:

    1 + cos2x
    = 1 + 2cos^2 - 1
    = 2cos^2x
    Looking good so far.

    On the right hand side I have:

    cotx sin2x
    = cosx/sinx (sin2x)
    = (cosx/sinx) (2sinx cosx / 1)
    = 2sinx 2cos^2x / sin x
    Where'd the extra 2 come from? It disappears later, but two wrongs don't make a right.

    = sinx 2cos^2x
    You didn't cancel the sin(x).

    But that's clearly not right....
    What does your proof look like now?
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  5. #5
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    I don't think I'm following you on the right hand side...

    If I'm multiplying the numerators, does cosx (2sinx cosx) = 2sinx cos^2. and then 2sinx/sinx would just leave the 2 somehow to get 2cos^2x?? I'm a little confused. I thought 2sinx/sinx would equal sinx
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  6. #6
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    Quote Originally Posted by starshine84 View Post
    I don't think I'm following you on the right hand side...

    If I'm multiplying the numerators, does cosx (2sinx cosx) = 2sinx cos^2.
    Well, cos(x) ( 2 sin(x) cos(x) ) = 2 sin(x) cos^2(x) is a bit clearer way to write it (I'm a stickler for parentheses around all function arguments), but I think you've got the idea there.

    and then 2sinx/sinx would just leave the 2 somehow to get 2cos^2x??
    Pretty much.

    I'm a little confused. I thought 2sinx/sinx would equal sinx
    It would equal 2, actually.

    You're almost there! I would just clean up your notation a little and present it better.
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  7. #7
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    Okay, I think I'm with you. Thank you again!! This unit is giving me a lot of trouble for some reason.
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  8. #8
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    You're welcome!
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