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Math Help - Solve using trig identities

  1. #1
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    Solve using trig identities

    sin(3t)=\frac{\sqrt3}{2}

    First off, let me just say that I know how to solve it by taking the arcsin and then dividing by 3, that isn't what I need to do.

    It needs to be like this:
    sin(2t+t)

    I know that:
    sin(2t)cos(t)+cos(2t)+sin(t)
    then:
    2sin(t)cos(t)cos(t)+(1-2sin^2(t))+sin(t)

    2sin(t)cos^2(t)+1-2sin^3(t)

    Then I get stuck. I'm trying to solve for t.
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  2. #2
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    Quote Originally Posted by inspire View Post
    [
    It needs to be like this:
    sin(2t+t)

    I know that:
    sin(2t)cos(t)+cos(2t)+sin(t)
    then:
    2sin(t)cos(t)cos(t)+(1-2sin^2(t))+sin(t)


    Do you mean? \sin(2t)\cos(t)+\cos(2t)\sin(t)

    Then

    2\sin(t)\cos^2(t)+(1-2\sin^2(t))\sin(t)

    2\sin(t)\cos^2(t)+\sin(t)-2\sin^3(t)
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  3. #3
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    Actually

    \sin{(2t + t)} = \sin{2t}\cos{t} + \cos{2t}\sin{t}

     = 2\sin{t}\cos{t}\cos{t} + (1 - 2\sin^2{t})\sin{t}

     = 2\sin{t}\cos^2{t} + \sin{t} - 2\sin^3{t}

     = 2\sin{t}(1 - \sin^2{t}) + \sin{t} - 2\sin^3{t}

     = 2\sin{t} - 2\sin^2{t} + \sin{t} - 2\sin^3{t}

     = 3\sin{t} - 4\sin^3{t}.


    So \displaystyle 3\sin{t} - 4\sin^3{t} = \frac{\sqrt{3}}{2}

    \displaystyle 0 = 4\sin^3{t} - 3\sin{t} + \frac{\sqrt{3}}{2}.


    Now try to solve.
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  4. #4
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    Quote Originally Posted by pickslides View Post
    Do you mean? \sin(2t)\cos(t)+\cos(2t)\sin(t)

    Then

    2\sin(t)\cos^2(t)+(1-2\sin^2(t))\sin(t)

    2\sin(t)\cos^2(t)+\sin(t)-2\sin^3(t)
    Isn't the addition for sin(\alpha+\beta)=sin\alpha cos\beta+cos\alpha sin\beta?
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  5. #5
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    yes, but that's not the form you used in your OP.

    Quote Originally Posted by inspire View Post
    sin(2t+t)

    I know that:
    sin(2t)cos(t)+cos(2t)+sin(t)
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  6. #6
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    Quote Originally Posted by Prove It View Post
    Actually

    \sin{(2t + t)} = \sin{2t}\cos{t} + \cos{2t}\sin{t}

     = 2\sin{t}\cos{t}\cos{t} + (1 - 2\sin^2{t})\sin{t}

     = 2\sin{t}\cos^2{t} + \sin{t} - 2\sin^3{t}

     = 2\sin{t}(1 - \sin^2{t}) + \sin{t} - 2\sin^3{t}

     = 2\sin{t} - 2\sin^2{t} + \sin{t} - 2\sin^3{t}

     = 3\sin{t} - 4\sin^3{t}.


    So \displaystyle 3\sin{t} - 4\sin^3{t} = \frac{\sqrt{3}}{2}

    \displaystyle 0 = 4\sin^3{t} - 3\sin{t} + \frac{\sqrt{3}}{2}.


    Now try to solve.
    Am I missing something, or is this incredibly hard to factor?
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