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Math Help - cos pi/6

  1. #1
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    cos pi/6

    quick question

    how is cos pi/6 = sq root 3 /2 I just can`t work it out

    DL
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  2. #2
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    Re: cos pi/6

    are you familiar w/ the side ratios of a 30-60-90 triangle? if so, let the hypotenuse have length = 1. also ...

    \cos(\theta) = \frac{opposite \, side}{adjacent \, side}
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  3. #3
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    Re: cos pi/6

    Hi Skeeter,

    a little correction:
    you did write tan(tau),

    cos(tau) = adjacent side / hypotenuse
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  4. #4
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    Re: cos pi/6

    yes, my error ... had tangent on the brain
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  5. #5
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    Re: cos pi/6

    Hello, davellew69!

    \text{How is: }\,\cos\tfrac{\pi}{6}\,=\,\frac{\sqrt{3}}{2}\,?

    Consider an equilateral triangle with side length 2.
    Each vertex has a 60o angle.
    Draw an altitude h. .It bisects the angle.
    The diagram looks like this.
    Code:
                  A
                  *
                 /|\
                / | \
               /30|  \
            2 /   |h  \ 2
             /    |    \
            /     |     \
           / 60   |      \
        B *-------*-------* C
              1   D   1
    Pythagorus says: . h^2 + 1^2 \,=\,2^2

    From which we get: . h \,=\,\sqrt{3}


    Hence, in a 30-60 right triangle, we have these sides:

    . . \begin{Bmatrix}\text{opposite }30^o & 1 \\ \text{opposite }60^o\!: & \sqrt{3} \\ \text{hypotenuse:} & 2 \end{Bmatrix}

    Now we can write all the trig values for 30o and 60o.

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