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Math Help - Another Pre-Calc Trig Problem

  1. #1
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    Another Pre-Calc Trig Problem

    given that:
    t=sin(18degrees)
    what are the steps in solving:
    (8t^4)-(8t^2)-t+1=0 ?
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  2. #2
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    Re: Another Pre-Calc Trig Problem

    Hello, jonathanraxa!

    I don't understand . . .


    Given that: . t \:=\:\sin(18^o)
    what are the steps in solving: . 8t^4 - 8t^2 -t+1\:=0\:\,?

    We don't have to solve the equation!

    They gave us the answer! . t \:=\:\sin18^o


    Can you give us the original wording of the problem?

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  3. #3
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    Re: Another Pre-Calc Trig Problem

    The original equation factors out to:

    (t-1)(2t+1)(4t^2+2t-1)=0

    Since we know t is sin(18), the first two factors will not equal zero, which leaves the final one.

    Put that into the quadratic formula and one of the roots you get is (-1+sqrt(5))/4


    (-1+sqrt(5))/4=sin(18)
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  4. #4
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    Re: Another Pre-Calc Trig Problem

    I'm plugging sin(18degrees) in for t in the equation.
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  5. #5
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    Re: Another Pre-Calc Trig Problem

    Guess I misunderstood the question.
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