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Math Help - Some trig problems from Shy Guy

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    Forum Admin topsquark's Avatar
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    Some trig problems from Shy Guy

    Quote Originally Posted by >_<SHY_GUY>_<
    one thing to just clear out...im kinda blank today...why is cosine of 0 = 1 and sine of 0 = 0?

    how is arctan square root of 3 divided by 3 = pi/6?

    how is (arctan x = pi/4) equal to 1?

    how is (arctan x = -pi/6) equal to the negative square root of 3 divided by 3?

    and how is square root 3 /3 part of the 30-60-90 triangle? i though it was sqaureroot 3/2?
    Imagine the unit circle. A point (x, y) has coordinates (cos(\theta), sin(\theta)). As the angle gets smaller and smaller the point on the unit circle gets closer and closer to (1, 0). Which means that cos(0) = 1 and sin(0) = 0.

    Think of the tangent problems backward.
    tan \left ( \frac{\pi}{6} \right ) = \frac{\sqrt{3}}{3}
    so
    tan^{-1} \left ( \frac{\sqrt{3}}{3} \right ) = \frac{\pi}{6}.

    The other inverse tangent problems follow a similar logic.

    The 30 - 60 - 90 triangle contains a 30 degree angle. tan(30) = sqrt(3) / 3. You are thinking of either sin(60) or cos(30), which are both sqrt(3) / 2.

    -Dan
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    Member >_<SHY_GUY>_<'s Avatar
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    i dont get how sqrt 3/3 in a 30-60-90 triangle in reference to the unit circle, is part of it...i dont know if that make sense...
    because i know that 1/2 and sqrt 3/2 and the hypotenuse 1 is a 30-60-90.

    and how do you know sqrt 3/3 is pi/6?
    Last edited by >_<SHY_GUY>_<; April 15th 2008 at 07:58 PM. Reason: i was missing one more question and didnt want to double post
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    i dont get how sqrt 3/3 in a 30-60-90 triangle in reference to the unit circle, is part of it...i dont know if that make sense...
    because i know that 1/2 and sqrt 3/2 and the hypotenuse 1 is a 30-60-90.

    and how do you know sqrt 3/3 is pi/6?
    For the tan(\pi / 6) part, this is just the same as tan(30), so let's work on that.

    tan(30) = \frac{sin(30)}{cos(30)} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}

    = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}

    Does this help?

    -Dan
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