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Math Help - Unit circle and exact values

  1. #1
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    Unit circle and exact values

    Hi all! I need serious help with unit circle and exact values! I never know how to do the symmetry. I was wondering whether i could get some help with this question here:

    I have to find the exact value:

    tan (-π/6)

    {In words: tan (negative pie divided by 6)}

    Is it possible to have it explained step by step or perhaps by using a unit circle?

    Help would be greatly appreciated =D
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  2. #2
    Member Glaysher's Avatar
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    Are you aware of the CAST diagram?
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  3. #3
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    If you imagine it like a clock, the 'beginning' of the unit circle is at 3 O'clock.

    pi is 180degrees, in otherwords 9 o'clock. You always move anticlockwise, unless it's a negative number. Ie: pi/2 = 12 o'clock and -pi/2 = 6 o'clock.

    If you divide the unit circle into four quadrants (3 - 12, 12 - 9, 9 - 6 and 6- 3) different functions are positive in different quadrants.

    Sine is positive in top left and top right
    Cosine is positive in top right and bottom right
    tangent is positive in top right and bottom left.

    It is also very handy to memorise few basic triangles. The most common two are:
    right angled, with two pi/4 angles.
    right angled with a pi/6 and pi/3 angle.

    Now we need to combine all of this knowledge. First work out tan(pi/6) and ignore the negative sign. From the triangles (the second one to be specific) this will yield  \frac{\sqrt{3}}{3} .

    Now that we have that exact value, we refer to the unit circle to see if it will be positive or negative.
    We move around the unit circle in the negative direction (because it's -pi/6) and we will not reach 6 o'clock (-pi/2). From this we can see that it's in the bottom right quadrant. This is where only cosine is positive. Therefore tangent will be negative.

    So our answer will be:  -\frac{\sqrt{3}}{3}

    PS: sorry for the long winded explanation if you already knew most of this, also, if you don't understand the triangles it's probably best that you refer to another source for this as it's very difficult to explain without a diagram.
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  4. #4
    Member Glaysher's Avatar
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    If you look at the diagram you see a CAST diagram with your angle of (-pi/6) illustrated on it. (Minus means measure clockwise ratther than anti-clockwise).

    The angle is in the C quadrant which means only cos is positive in this quadrant and therefore sin and tan give negative values.

    If you work out tan (pi/6) then either tan (-pi/6) = tan (pi/6) or tan (-pi/6) = -tan (pi/6)

    pi/6 is in the A quadrant which stands for ALL meaning that sin, cos and tan are all positive in that quadrant.

    But tan is negative in the C quadrant and positive in the A quadrant so this means tan (-pi/6) = -tan (pi/6)
    Attached Thumbnails Attached Thumbnails Unit circle and exact values-cast.png  
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  5. #5
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    CAST

    Quote Originally Posted by Glaysher View Post
    Are you aware of the CAST diagram?
    Yes i am aware of CAST and thank you for the explanation, it has really clarified a few things

    However,im unsure about how pi/6 ends up in the 4th quadrant and the how the symmetry of it works. Is there like an equivalent angle?

    Cheers =D
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  6. #6
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    Quote Originally Posted by steph_r View Post
    Yes i am aware of CAST and thank you for the explanation, it has really clarified a few things

    However,im unsure about how pi/6 ends up in the 4th quadrant and the how the symmetry of it works. Is there like an equivalent angle?

    Cheers =D
    The symmetry works in all four directions, ie; ignoring the positive or negative signs,  sin(\frac{\pi}{6}) = sin(\frac{5\pi}{6}) = sin(\frac{7\pi}{6}) = sin(\frac{11\pi}{6})
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  7. #7
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    exact values

    Quote Originally Posted by U-God View Post
    The symmetry works in all four directions, ie; ignoring the positive or negative signs,  sin(\frac{\pi}{6}) = sin(\frac{5\pi}{6}) = sin(\frac{7\pi}{6}) = sin(\frac{11\pi}{6})
    So am i right to say that:

    cos (-pi/6) = (-root3/2)
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  8. #8
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    Quote Originally Posted by steph_r View Post
    So am i right to say that:

    cos (-pi/6) = (-root3/2)
    No, Cosine is positive in the bottom right quadrant
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