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Math Help - Trig issues...

  1. #1
    Pyotr
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    Question Trig issues...

    I am completely and utterly stuck on this one problem:



    I am supposed to verify this identity using only the fundamental identities. Could somebody at least lead me in the right way to start this? I have no clue where to even begin with this one
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Pyotr
    I am completely and utterly stuck on this one problem:



    I am supposed to verify this identity using only the fundamental identities. Could somebody at least lead me in the right way to start this? I have no clue where to even begin with this one
    Here we go:

    <br />
\frac{\sin(x)}{1+\cos(x)}+\frac{1+\cos(x)}{\sin(x)  }=\frac{(\sin(x))^2+(1+cos(x))^2}{(1+\cos(x))\sin(  x)}<br />
    <br />
=\frac{(\sin(x))^2+1+2\cos(x)+(cos(x))^2}{(1+\cos(  x))\sin(x)}= \frac{\{(\sin(x))^2+(cos(x))^2\}+1+2\cos(x)}{(1+\c  os(x))\sin(x)}<br />

    <br />
=\frac{2+2\cos(x)}{(1+\cos(x))\sin(x)}=\frac{2}{ \sin (x)}<br />
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  3. #3
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    I did it this way:

    \frac{\sin(x)}{1+\cos(x)}+\frac{1+\cos(x)}{\sin(x)  }=\frac{2}{\sin(x)}

    \frac{\sin^2(x)+(1+\cos(x))^2}{\sin(x)(1 + \cos(x))} = \frac{2}{\sin(x)}

    \sin^2(x)+(1+\cos(x))^2 - 2\sin(x)(1+\cos(x)) = 0

    (\sin(x) - (1+\cos(x)))^2 = 0

    (\sin(x) - \cos(x) - 1)^2 = 0
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by chancey
    I did it this way:

    \frac{\sin(x)}{1+\cos(x)}+\frac{1+\cos(x)}{\sin(x)  }=\frac{2}{\sin(x)}

    \frac{\sin^2(x)+(1+\cos(x))^2}{\sin(x)(1 + \cos(x))} = \frac{2}{\sin(x)}

    \sin^2(x)+(1+\cos(x))^2 - 2\sin(x)(1+\cos(x)) = 0

    (\sin(x) - (1+\cos(x)))^2 = 0

    (\sin(x) - \cos(x) - 1)^2 = 0
    Your third line does not follow from your second.

    RonL
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by chancey
    I did it this way:

    \frac{\sin(x)}{1+\cos(x)}+\frac{1+\cos(x)}{\sin(x)  }=\frac{2}{\sin(x)}

    \frac{\sin^2(x)+(1+\cos(x))^2}{\sin(x)(1 + \cos(x))} = \frac{2}{\sin(x)}

    \sin^2(x)+(1+\cos(x))^2 - 2\sin(x)(1+\cos(x)) = 0

    (\sin(x) - (1+\cos(x)))^2 = 0

    (\sin(x) - \cos(x) - 1)^2 = 0
    Even if this were correct it would not as it stands constitute a proof that
    the first line is an identity, because if this were correctly done you will
    have shown that if you assume the first line is an identity then the last
    line is an identity (which if you had done this right it of course would have
    been).

    You will need to work the other way around. Start with a known identity
    and from that derive the equation you wish to show is an identity.

    RonL
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  6. #6
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    Woops, thanks for pointing that out, I didn't to multiply the other side of the fraction, I thought the answer was strange

    \frac{\sin^2(x)+(1+\cos(x))^2}{\sin(x)(1+\cos(x))} = \frac{2}{\sin(x)}

    \sin(x)(\sin^2(x)+(1+\cos(x))^2) - 2\sin(x)(1+\cos(x)) = 0

    \sin^2(x)+(1+\cos(x))^2 - 2(1+\cos(x)) = 0

    \sin^2(x)+1+2\cos(x)+\cos^2(x) - 2-2\cos(x) = 0

    1+1+2\cos(x) - 2-2\cos(x) = 0

    0 = 0
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by chancey
    Woops, thanks for pointing that out, I didn't to multiply the other side of the fraction, I thought the answer was strange

    \frac{\sin^2(x)+(1+\cos(x))^2}{\sin(x)(1+\cos(x))} = \frac{2}{\sin(x)}

    \sin(x)(\sin^2(x)+(1+\cos(x))^2) - 2\sin(x)(1+\cos(x)) = 0

    \sin^2(x)+(1+\cos(x))^2 - 2(1+\cos(x)) = 0

    \sin^2(x)+1+2\cos(x)+\cos^2(x) - 2-2\cos(x) = 0

    1+1+2\cos(x) - 2-2\cos(x) = 0

    0 = 0
    My other comment still applies.

    RonL
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  8. #8
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    Quote Originally Posted by CaptainBlack
    My other comment still applies.
    I don't understand what you mean? The steps show that the equation is correct and I did use one of the trig identities to simplify...

    EDIT: Ohhh I just understood what you meant as I clicked the button, you mean I need to manipulate the LHS to equal the RHS to show that they are equal?
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by chancey
    I don't understand what you mean? The steps show that the equation is correct and I did use one of the trig identities to simplify...

    EDIT: Ohhh I just understood what you meant as I clicked the button, you mean I need to manipulate the LHS to equal the RHS to show that they are equal?
    Exactly.

    RonL
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