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Math Help - solving this identity

  1. #1
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    solving this identity

    (sin (2x) +sin(4x)/sin(2x)-sin(4x)) + (tan(3x))/tanx)

    i'm totally lost o.o
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  2. #2
    Super Member bigwave's Avatar
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    Cool

    first, this is just an expression
    not an identity we can prove
    there is no = sign

    otherwise ..

    <br />
\sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\c  os\frac{\alpha-\beta}{2}<br />

    and

    <br />
\sin\alpha-\sin\beta=2\cos\frac{\alpha+\beta}{2}\sin\frac{\al  pha-\beta}{2}<br />
    Last edited by bigwave; December 28th 2009 at 08:50 PM.
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  3. #3
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    Quote Originally Posted by alessandromangione View Post
    (sin (2x) +sin(4x)/sin(2x)-sin(4x)) + (tan(3x))/tanx)

    i'm totally lost o.o
    1. There's no such thing as "solving an identity". To solve means to find the value of x (or whatever the unknown is) that satisfies an equation. An identity is an equation that is true no matter what the x value.

    2. You can PROVE an identity - i.e. show that two expressions are identically equal. Is that what you are trying to do? What are you trying to show that this is the same as? Or are you trying to ESTABLISH an identity? What was said above - there is not an equals sign so this is not an identity.
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  4. #4
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    guys i'm really sorry...i forgot to put '=0'...i'm sorry
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  5. #5
    Super Member bigwave's Avatar
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    Cool first step

    given identity prove that:

    <br />
\frac{\sin{2x} +\sin{4x}}{\sin{2x}-\sin{4x}} + \frac{\tan{3x}}{\tan{x}}=0

    use sum to product formulas

    \frac<br />
{2\sin\frac{2x+4x}{2}\cos\frac{2x-4x}{2}}<br />
{2\cos\frac{2x+4x}{2}\sin\frac{2x-4x}{2}}<br />
+\frac{\tan{3x}}{tan{x}}=0<br />

    can you finish from here?
    Last edited by bigwave; December 28th 2009 at 10:02 PM. Reason: wording
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  6. #6
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    thanks
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