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Math Help - Exact Value

  1. #1
    Senior Member DivideBy0's Avatar
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    Exact Value

    I have to evaluate the exact value of tan(15).

    To give some context, I have to find the area of a dodecagon, and I used the uber super cool formula:
    ns^2/(4tan(180/n)) where n is the number of sides and s is the side length.

    Thanks.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    I have to evaluate the exact value of tan(15).

    To give some context, I have to find the area of a dodecagon, and I used the uber super cool formula:
    ns^2/(4tan(180/n)) where n is the number of sides and s is the side length.

    Thanks.
    Use a half-angle formula.

    I can never remember the formula for tan(a/2), so I do it this way:

    sin(a/2) = (+/-)sqrt{1 - cos(a)}/sqrt{2}
    cos(a/2) = (+/-)sqrt{1 + cos(a)}/sqrt{2}
    (where we get the + or - from which quadrant the angle is in.)

    So
    tan(a/2) = sin(a/2)/cos(a/2) = (+/-)sqrt{1 - cos(a)}/sqrt{1 + cos(a)}

    In this case, a = 30:
    tan(15) = sqrt{1 - cos(30)}/sqrt{1 + cos(30)}

    = sqrt{1 - sqrt{3}/2}/sqrt{1 + sqrt{3}/2}

    = sqrt{2 - sqrt{3}}/sqrt{2 + sqrt{3}}

    You will want to rationalize this, so multiply the numerator and denominator by sqrt{2 - sqrt{3}}. I'll just give you the answer:

    tan(15) = sqrt{7 - 4sqrt{3}}

    -Dan
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  3. #3
    Senior Member DivideBy0's Avatar
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    Thanks m8!
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    Thanks m8!
    I appreciate the thanks ( ), but what the heck is "m8?"

    -Dan
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  5. #5
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by topsquark View Post
    I appreciate the thanks ( ), but what the heck is "m8?"

    -Dan
    It's a 'leet' abbreviation of 'mate' (used commonly in australia)
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    It's a 'leet' abbreviation of 'mate' (used commonly in australia)
    Ah! Obvious once it's explained.

    -Dan
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