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Math Help - Trig identities help

  1. #1
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    Trig identities help

    Hi i have question

    tanx - cotx / tanx + cotx + 1 = 2sinx

    so...
    sin/cos - cos/sin / sin/cos + cos/sin

    then mutlple fracs and get

    sin^2x - cos ^2x/ sinxcosx / sin^2x + cos^2x/ sinxcosx

    and then what do i do?
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  2. #2
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    Is this your question? \frac{\tan{x}-\cot{x}}{\tan{x}+\cot{x}} + 1 = 2\sin{x}

    I'm going to assume that the left hand side is written correctly, but I'm going to ask you to recheck the right hand side.

    After this step:
    \frac{\frac{\sin{x}}{\cos{x}} - \frac{\cos{x}}{\sin{x}}}{\frac{\sin{x}}{\cos{x}} + \frac{\cos{x}}{\sin{x}}} \cdot \frac{\sin{x}\cos{x}}{\sin{x}\cos{x}}

    We get:
    \frac{\sin^2{x}-\cos^2{x}}{\sin^2{x}+\cos^2{x}} + 1

    Recall that \sin^2{x}+\cos^2{x} = 1

    Therefore:

    = \sin^2{x}-\cos^2{x}+1 = 2\sin^2{x}
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  3. #3
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    Quote Originally Posted by Chop Suey View Post
    Is this your question? \frac{\tan{x}-\cot{x}}{\tan{x}+\cot{x}} + 1 = 2\sin{x}

    I'm going to assume that the left hand side is written correctly, but I'm going to ask you to recheck the right hand side.

    After this step:
    \frac{\frac{\sin{x}}{\cos{x}} - \frac{\cos{x}}{\sin{x}}}{\frac{\sin{x}}{\cos{x}} + \frac{\cos{x}}{\sin{x}}} \cdot \frac{\sin{x}\cos{x}}{\sin{x}\cos{x}}

    We get:
    \frac{\sin^2{x}-\cos^2{x}}{\sin^2{x}+\cos^2{x}} + 1

    Recall that \sin^2{x}+\cos^2{x} = 1

    Therefore:

    = \sin^2{x}-\cos^2{x}+1 = 2\sin^2{x}
    Sorry it is 2sin^2x, but i still dont understand the steps
    i see how you got
    \frac{\sin^2{x}-\cos^2{x}}{\sin^2{x}+\cos^2{x}} + 1
    but after that how did you end up with 2sin^2? how do you divide that :S
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  4. #4
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    Quote Originally Posted by cokeclassic View Post
    how do you divide that :S
    I didn't. I used the Pythagorus identity \sin^2{x}+\cos^2{x} = 1

    Then notice that 1-\cos^2{x} = \sin^2{x}
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  5. #5
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    Ok I see.... thanks...

    so basically

    sin^2 + cos^2/ sin^2 + cos^2 is the same as
    sin^2 + cost ^2 since it equal 1
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  6. #6
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    \frac{\sin^2{x} + \cos^2{x}}{\sin^2{x} + \cos^2{x}} = 1

    But I think you might have made a typo and meant instead:
    \frac{\sin^2{x} - \cos^2{x}}{\sin^2{x} + \cos^2{x}}

    In that case, yes, it equals \sin^2{x} - \cos^2{x} since \sin^2{x} + \cos^2{x} = 1
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  7. #7
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    Er yeah, ty
    Last edited by cokeclassic; November 9th 2008 at 07:40 PM. Reason: mistake
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