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Math Help - Problem Need Explatation: Finding an algebraic number..

  1. #1
    Newbie KidBlondie's Avatar
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    Problem Need Explatation: Finding an algebraic number..

    I知 not sure how to arrive at these answers. For instance, how do I know when a sin or tan is negative? I tried to draw a diagram next to the question to help me understand but it didn稚 work out to well. If someone could explain this for me it would be greatly appreciated.
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  2. #2
    Member Goku's Avatar
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    Re: Problem Need Explatation: Finding an algebraic number..

    1st quadrant - All are positive only
    2nd quadrant - sine is positive only
    3rd quadrant - tan is positive only
    4th quadrant - cos is positive only

    Why?
    cos(theta) = x/r and since x is positive implies x/r is positive... and the saame for the rest.
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  3. #3
    Newbie KidBlondie's Avatar
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    Re: Problem Need Explatation: Finding an algebraic number..

    Okay, so where can I find the proof of this?
    And how are the algebraic numbers being determined? If my graph of the reference angle is correct, sine should be opposite/adjacent. I知 not seeing where the ス or any of the other algebraic numbers are coming from.
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  4. #4
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    e^(i*pi)'s Avatar
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    Re: Problem Need Explatation: Finding an algebraic number..

    Quote Originally Posted by KidBlondie View Post
    Okay, so where can I find the proof of this?
    And how are the algebraic numbers being determined? If my graph of the reference angle is correct, sine should be opposite/adjacent. I知 not seeing where the ス or any of the other algebraic numbers are coming from.
    Proof comes from the definitions of the trig functions from a right angled triangle which is effectively superimposed onto the unit circle where the angle is always measured from the origin. The hypotenuse (r) is always positive because it is defined via Pythagoras r =\sqrt{x^2+y^2} for the unit circle and a property of a square number is that it's non-negative for all real numbers. In other words r is taken to be positive regardless of it's direction so it doesn't affect the sign in the quadrants

    On the unit circle the opposite is parallel to the y-axis (check on your diagram) and the adjacent to the x-axis

    \sin(x) = \dfrac{opp}{hyp} = \dfrac{y}{|r|}

    \cos(x) = \dfrac{adj}{hyp} = \dfrac{x}{|r|}

    If sin and cos have the same sign then tan is positive, if not then tan is negative
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  5. #5
    Newbie KidBlondie's Avatar
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    Re: Problem Need Explatation: Finding an algebraic number..

    I understand the positive and negative values for the trig functions now. Thank you both. But how is the the cos ス?
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  6. #6
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    Re: Problem Need Explatation: Finding an algebraic number..

    Quote Originally Posted by KidBlondie View Post
    I understand the positive and negative values for the trig functions now. Thank you both. But how is cos \theta= ス?
    The right-angled triangle side relationships
    gives the first quadrant situation.

    When your angle is in any of the other 3 quadrants,
    the method I use is, for any point (x,y) on the circle circumference....

    cos\theta=\frac{"x"\;co-ord}{R}

    sin\theta=\frac{"y"\;co-ord}{R}

    tan\theta= the slope of the line going through the origin and the point (x,y)

    Note that the rules for whether sine, cosine or tangent are positive or negative
    concur with this simple view.

    Hence, in your example

    cos(2\pi-\theta)=cos\theta

    due to axial symmetry in the horizontal axis.
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  7. #7
    Newbie KidBlondie's Avatar
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    Re: Problem Need Explatation: Finding an algebraic number..

    But is knowing the reference angle enough information to deduce the point of intersection between ray and the circle? Can this be done without that information?

    Quote Originally Posted by Archie Meade View Post
    Hence, in your example

    cos(2\pi-\theta)=cos\theta
    How do I find the 度 co-cord and 土 co-cord with the information given?
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  8. #8
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    Re: Problem Need Explatation: Finding an algebraic number..

    The co-ordinates will tell you whether cosine is positive or negative.
    You don't need their values.

    "x" is positive in the 4th quadrant, so cosine is positive there,
    and the cosine of the angle in the 4th quadrant is the exact same as the cosine of 60 degrees,
    the reference angle.
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  9. #9
    Newbie KidBlondie's Avatar
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    Re: Problem Need Explatation: Finding an algebraic number..

    Yes, I'm still with you the: 60 degrees is equivalent to pi/3. But forgive me, how do I find the cos of this?
    Quote Originally Posted by Archie Meade View Post
    Without knowing the x-cord or the radius..
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  10. #10
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    Re: Problem Need Explatation: Finding an algebraic number..

    Quote Originally Posted by KidBlondie View Post
    Yes, I'm still with you the: 60 degrees is equivalent to pi/3. But forgive me, how do I find the cos of this? Without knowing the x-cord or the radius..
    You ought to just use a table of common reference angles
    from a book of mathematics tables.
    There are standard "surd" solutions for 30, 45, 60 degrees.

    I added my comments to help with understanding why sine, cosine and tangent
    are positive and negative in the various quadrants.

    If you look at the right-angled triangle in your first post,
    the ratio of the base (on the horizontal x-axis) to the hypotenuse is 1:2,
    and since "x" is positive in the 1st and 4th quadrants, cosine is positive in those quadrants.
    Looking at where the triangle intersects the circle circumference,
    if you draw the same triangle in the 1st quadrant, the base resting on the x-axis
    and the point of contact with the circle being directly above that for the 4th quadrant triangle,
    then you see that cosine is the same for both, because the "x" co-ordinate is the same.
    Last edited by Archie Meade; July 8th 2011 at 04:30 PM.
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  11. #11
    Newbie KidBlondie's Avatar
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    Re: Problem Need Explatation: Finding an algebraic number..

    Indeed. I found the information I needed by looking up the term "table of common reference angles". Thanks.
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