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Math Help - Find>trigo~!

  1. #1
    Junior Member
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    Post Find>trigo~!

    Tan3A=cot(∏/3-A), for 0≤A≤360

    Tan3A =????
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  2. #2
    MHF Contributor
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    Here is one way. The complementary angles way.

    sinA = cos(pi/2 -A)
    Likewise, tanA = cot(pi/2 -A)
    So,
    tan(3A) = cot(pi/3 -A)
    Transform the LHS side into its cot form,
    cot(pi/2 -3A) = cot(pi/3 -A)
    So,
    pi/2 -3A = pi/3 -A
    pi/2 - pi/3 = -A +3A
    pi/6 = 2A
    A = pi/12 ------------answer.
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  3. #3
    Super Member

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    Lexington, MA (USA)
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    Hello, sanikui!

    This could be an ugly problem . . .


    \tan 3x\:=\:\cot(60^o -x),\;\text{ for }0^o\,\leq\,x \leq \,360^o

    \text{Evaluate: }\tan3x
    There is a back-door approach to this problem
    . . if we consider acute angles only.


    If \tan A \:=\:\cot B. then A and B are complements.

    Look at this triangle:
    Code:
                            *
                         * B|
                      *     |
                   *        | y
                *           |
             * A            |
          * - - - - - - - - *
                   x

    We have: . \tan A \:=\:\frac{y}{x},\;\;\cot B \:=\:\frac{y}{x}

    Since A and B are in the same right triangle:. A + B \:=\:90^o


    So we have: . 3x + (60^o - x) \;=\;90^o\quad\Rightarrow\quad 2x \:=\:30^o \quad\Rightarrow\quad x \:=\:15^o<br />


    Therefore: . \tan(3x) \;=\;\tan45^o \;=\;1

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