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    I am lost on this. Can someone help:

    1-2cos(squared)x = sinx-cosx
    _______________ ________

    1-2cosxsinx sinx-cosx



    you either have to make them equal to eachother or one equal to the other. that line is the divide line. and I couldnt find a better way to type squared but theres no parentheses there, just for clarification.

    HELP!!!! Thanks

    Josh

    and the first bottm equation is: 1-2cosxsinx
    the second bottom is: sinx-cosx

    hope this helps
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    No my nightmares have returned! Trig Identities...noooooooooooooooooo oooooooooo :eek:
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    I'm sorry, but what do you want to prove? I can't understand your formatting.
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    (Original post by theone)
    I'm sorry, but what do you want to prove? I can't understand your formatting.
    (1-2cos²x)/(1-2 cosx sinx) == (sinx - cosx)/(sinx - cosx) [===1 ????? :confused: ]
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    (Original post by elpaw)
    (1-2cos²x)/(1-2 cosx sinx) == (sinx - cosx)/(sinx - cosx) [===1 ?????]

    But this isn't true.
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    (Original post by theone)
    But this isn't true.
    which is why i have put all them question marks
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    (Original post by elpaw)
    which is why i have put all them question marks
    Perhaps he means (1-2cos²x)/(1-2 cosx sinx) == (sinx + cosx)/(sinx - cosx), which I think could be true.

    oops....yes your right.....


    (1-2cos²x)/(1-2 cosx sinx) == (sinx + cosx)/(sinx - cosx)


    thats it

    I think its a sum and difference identity and the formula is either:

    sin(A+B) = sinA cosB + cosA sinB

    or

    sin(A-B) = sinA cosB - cosA sinB

    I am completely lost though and theres another one if your up for it, but that one will end in a fraction.....GULP....

    josh
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    (Original post by Unregistered)

    I am completely lost though and theres another one if your up for it

    josh
    Sure.
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    And your previous one the easiest way to go seems to be to conside the identities for Sin A - / + Cos B
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    1 = sin²x + cos²x

    => 1 - 2 cos²x == sin²x - cos²x == (sinx + cos x)(sin x - cos x)
    ----------------------------------------------------------------------
    1 - 2 cosx sinx == sin²x + cos²x - 2 sinx cosx == (sinx - cosx)²

    (sinx - cosx) cancells from top and bottom

    => == (sinx + cosx)/(sinx - cosx)
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    Or that which is a far better method.

    heres the other problem:


    tan(x-y) -tan(y-x) == 2(tanx-tany)/1+tanx tany

    I know this ends in a fraction, and it doesnt seem as hard as the other problem. I know the formula is either:

    tan(A+B) == tanA+tanB/1-tanA tanB

    or

    tan(A-B) == tanA-tanB/1+tanA tanB

    thanks guys

    josh

    it looks like you may have to work out both sides of equation to make them equal eachother. My brain hurts just thinking about it....
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    (Original post by Unregistered)
    heres the other problem:


    tan(x-y) -tan(y-x) == 2(tanx-tany)/1+tanx tany

    I know this ends in a fraction, and it doesnt seem as hard as the other problem. I know the formula is either:

    tan(A+B) == tanA+tanB/1-tanA tanB

    or

    tan(A-B) == tanA-tanB/1+tanA tanB

    thanks guys

    josh
    tan(x-y) - tan(y-x) == ((tanx - tany) - (tany - tanx))/(1+ tanx tany)

    == 2(tanx - tany)/(1+tanx tany)

    Simple... just use the identities, like you said....

    oh duh....I knew it was easy.

    But thanks again....I bow down to almighty identity god

    josh
 
 
 
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