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Math Help - Show me where I went wrong, trig equation

  1. #1
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    Show me where I went wrong, trig equation

     cos\theta = cos(\theta+60)

     cos\theta = cos\theta cos60 - sin\theta sin60

     cos\theta = \frac{1}{2}cos\theta-\frac{\sqrt{3}}{2} sin\theta

    than multiply eachterm by '2', so I get;

     2cos\theta = cos\theta - \sqrt{3}sin\theta

     2cos\theta - cos\theta = - \sqrt{3}sin\theta

     cos\theta = - \sqrt{3}sin\theta

     \frac{cos\theta}{sin\theta} = -\sqrt{3}


     tan\theta = -\sqrt{3}

    my book says the answer is  tan\theta = -\frac{1}{\sqrt{3}}

    Can someone please show me what I have done wrong, cause I have been through my working quite a few times still get the same answer!

    Thank you
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  2. #2
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    Quote Originally Posted by Tweety View Post
     cos\theta = cos(\theta+60)

     cos\theta = cos\theta cos60 - sin\theta sin60

     cos\theta = \frac{1}{2}cos\theta-\frac{\sqrt{3}}{2} sin\theta

    than multiply eachterm by '2', so I get;

     2cos\theta = cos\theta - \sqrt{3}sin\theta

     2cos\theta - cos\theta = - \sqrt{3}sin\theta

     cos\theta = - \sqrt{3}sin\theta

     \frac{cos\theta}{sin\theta} = -\sqrt{3}


     tan\theta = -\sqrt{3}

    my book says the answer is  tan\theta = -\frac{1}{\sqrt{3}}

    Can someone please show me what I have done wrong, cause I have been through my working quite a few times still get the same answer!

    Thank you
    Only your last step is wrong

    \frac{cos(x)}{sin(x)} \neq tan(x)

    \frac{cos(x)}{sin(x)} = cot(x) = \frac{1}{tan(x)}

    If you take the reciprocal of the answer you have it will be right
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  3. #3
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    yes the last step cos / sin isnot tan ,, sin / cos is tan so it will be ( cos / sin ) = - square root ( 3 )
    => 1/tan = -square root (3) => tan = 1/-squr root 3
    got it ?
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  4. #4
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    AAH thanks, can't actually believe that I did not spot that myself!

    I know that cos/sin does not equal tan, but was not looking at it properly.
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  5. #5
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    it happens tweety
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