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Core 2 ONE mark question... Driving me insane :|

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    Sin pheta + cos pheta = 0
    Show that tan pheta = -1

    Can someone quickly solve this, so I can move on with my life and look beyond the 30 minutes I've wasted...

    And er, thanks in advance.
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    remember that sin/cos=tan...
  3. Offline

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    (Original post by Contrad!ction.)
    remember that sin/cos=tan...
    this.


    Most likely what' going wrong is that you've forgotten that cancelling out by dividing by something leaves a 1 not a 0. I'm guessing you're arriving at tan(theta) = 0 ?

    it's a mistake that I admittedly made quite a few silly times at the start of my maths A-level
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    isn't it theta
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    (Original post by MSI_10)
    Sin pheta + cos pheta = 0
    Show that tan pheta = -1

    Can someone quickly solve this, so I can move on with my life and look beyond the 30 minutes I've wasted...

    And er, thanks in advance.
    remember that sin/cos = tan

    so use that fact to move the cos pheta across to give

    sin = - cos

    divide through by -cos

    -tan = 1

    therefore multiply by -1 to get tan as positive. simples..
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    sin + cos = 0
    sin = -cos
    sin/cos = -1
    tan = -1
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    Hmm thanks all

    Is it also possible to multiply the starting equation -1, then end up with tan theta=-1?
    sin theta + cos theta = 0

    -sin theta - cos theta = 0
    -sin theta=1+cos theta
    -sin theta/ cos theta = 1
    -tan theta=1

    tan theta=-1

    After posting, I did that and got it right.
    Unless it was a fluke
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    (Original post by MSI_10)
    Hmm thanks all

    Is it also possible to multiply the starting equation -1, then end up with tan theta=-1?
    sin theta + cos theta = 0

    -sin theta - cos theta = 0
    -sin theta=1+cos theta
    -sin theta/ cos theta = 1
    -tan theta=1

    tan theta=-1

    After posting, I did that and got it right.
    Unless it was a fluke
    Where did the 1 come from in the second line? It shouldn't be there. You should just have -sin theta=cos theta
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    (Original post by Gemini92)
    Where did the 1 come from in the second line? It shouldn't be there. You should just have -sin theta=cos theta
    Oh okay thanks.

    So if I didn't make that mistake, it would still work right?

    So to summarize, my mistake was that when doing sin theta / cos theta, the right hand side gets a +1 NOT a 0?
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    (Original post by MSI_10)
    Oh okay thanks.

    So if I didn't make that mistake, it would still work right?

    So to summarize, my mistake was that when doing sin theta / cos theta, the right hand side gets a +1 NOT a 0?
    Yes.

    If you divide cos theta by cos theta you get one

    Same as if you divide anything most things by themselves
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    (Original post by wibletg)
    Yes.

    If you divide cos theta by cos theta you get one

    Same as if you divide anything by itself
    0/0 doesn't equal 1.
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    (Original post by Mr M)
    0/0 doesn't equal 1.
    haha, fair point
  13. Offline

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    (Original post by Mr M)
    0/0 doesn't equal 1.
    I was tempted to put most things, infinity divided by infinity isn't one either
  14. Offline

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    (Original post by IamBeowulf)
    isn't it theta
    Nah, it's a new Greek letter this.

     \sin(\cjRL{\varphi\put(0,0){\put(0,0)  { \put(-5,0){\theta}}}}) + \cos(\cjRL{\varphi\put(0,0){\put(0,0)  { \put(-5,0){\theta}}}}) = 0

    Pheta. That's how it's made: \ \ \varphi \theta\ \rightarrow\ \cjRL{\varphi\put(0,0){\put(0,0)  { \put(-1,0){\theta}}}}\ \rightarrow\ \cjRL{\varphi\put(0,0){\put(0,0)  { \put(-2,0){\theta}}}}\ \rightarrow\ \cjRL{\varphi\put(0,0){\put(0,0)  { \put(-3,0){\theta}}}}\ \rightarrow\ \cjRL{\varphi\put(0,0){\put(0,0)  { \put(-4,0){\theta}}}}\ \rightarrow\ \cjRL{\varphi\put(0,0){\put(0,0)  { \put(-5,0){\theta}}}}
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    (Original post by gff)
    Nah, it's a new Greek letter this.

     \sin(\cjRL{\varphi\put(0,0){\put(0,0)  { \put(-5,0){\theta}}}}) + \cos(\cjRL{\varphi\put(0,0){\put(0,0)  { \put(-5,0){\theta}}}}) = 0

    Pheta. That's how it's made: \ \ \varphi \theta\ \rightarrow\ \cjRL{\varphi\put(0,0){\put(0,0)  { \put(-1,0){\theta}}}}\ \rightarrow\ \cjRL{\varphi\put(0,0){\put(0,0)  { \put(-2,0){\theta}}}}\ \rightarrow\ \cjRL{\varphi\put(0,0){\put(0,0)  { \put(-3,0){\theta}}}}\ \rightarrow\ \cjRL{\varphi\put(0,0){\put(0,0)  { \put(-4,0){\theta}}}}\ \rightarrow\ \cjRL{\varphi\put(0,0){\put(0,0)  { \put(-5,0){\theta}}}}
    That's cool. Unfortunately PRSOM.
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    (Original post by wibletg)
    I was tempted to put most things, infinity divided by infinity isn't one either
    0/0 isn't infinity, it's solution set is.
  17. Offline

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    (Original post by 122025278)
    0/0 isn't infinity, it's solution set is.
    I don't think he suggested it was.

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