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Math Help - Solve for t

  1. #1
    Super Member sakonpure6's Avatar
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    Solve for t

    I have the following equation y=-20cos225t+50 If i set y= 60 and i solve i get root 2 over 2, howevef if i sub that back in the equation, i don't get 60 but I get about 68! Any help is appreciated. Here is what i did:

    60=-20cos225t+50
    t=10/-20cos225
    t= root 2/2
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  2. #2
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    Re: Solve for t

    is the equation ...

    y = -20\cos(225t) + 50

    or

    y = -20\cos(225) \cdot t + 50
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  3. #3
    Super Member sakonpure6's Avatar
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    Re: Solve for t

    Quote Originally Posted by skeeter View Post
    is the equation ...

    y = -20\cos(225t) + 50

    or

    y = -20\cos(225) \cdot t + 50

    This one y = -20\cos(225t) + 50

    if it's the other equation then root2/2 works... so what did i do wrong?
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  4. #4
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    Re: Solve for t

    Quote Originally Posted by sakonpure6 View Post
    This one y = -20\cos(225t) + 50

    if it's the other equation then root2/2 works... so what did i do wrong?
    the whole thing ...

    60 = -20\cos(225t) + 50

    -10 = -20\cos(225t)

    \frac{1}{2} = \cos(225t)

    \cos^{-1}\left(\frac{1}{2}\right) = 225t

    if t is in radians ...

    \frac{\pi}{3} = 225t

    t = \frac{\pi}{675}

    if t is in degrees ...

    60 = 225t

    t = \frac{60}{225} = \frac{4}{15}

    note that these are the principle solutions ... there are more.
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  5. #5
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    Re: Solve for t

    Hello, sakonpure6!

    I don't understand your difficulty.


    y\:=\:-20(\cos225)t+50
    If I set y= 60 and I solve, I get \tfrac{\sqrt{2}}{2}
    However if I sub that back in the equation, I don't get 60 but I get about 68! . How?

    If t = \tfrac{\sqrt{2}}{2}, we have:

    . . \text{-}20(\cos225)\left(\tfrac{\sqrt{2}}{2}\right) + 50 \;=\;\text{-}20\left(\text{-}\tfrac{\sqrt{2}}{2}\right)\left(\tfrac{\sqrt{2}}{  2}\right) + 50

    . . . . =\;\text{-}20\left(\text{-}\tfrac{1}{2}\right) + 50 \;=\;10 + 50 \;=\;60
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  6. #6
    Super Member sakonpure6's Avatar
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    Re: Solve for t

    Quote Originally Posted by Soroban View Post
    Hello, sakonpure6!

    I don't understand your difficulty.



    If t = \tfrac{\sqrt{2}}{2}, we have:

    . . \text{-}20(\cos225)\left(\tfrac{\sqrt{2}}{2}\right) + 50 \;=\;\text{-}20\left(\text{-}\tfrac{\sqrt{2}}{2}\right)\left(\tfrac{\sqrt{2}}{  2}\right) + 50

    . . . . =\;\text{-}20\left(\text{-}\tfrac{1}{2}\right) + 50 \;=\;10 + 50 \;=\;60
    I multiplied root2/2 by 225 before taking it's cosine, that was my mistake there and for skeeter how did you get -10? shouldnt it be 10
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  7. #7
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    Re: Solve for t

    my mistake ...

    60 = -20\cos(225t) + 50

    10 = -20\cos(225t)

    -\frac{1}{2} = \cos(225t)

    \cos^{-1}\left(-\frac{1}{2}\right) = 225t

    if t is in radians ...

    \frac{2\pi}{3} = 225t

    t = \frac{2\pi}{675}

    if t is in degrees ...

    120 = 225t

    t = \frac{120}{225} = \frac{8}{15}
    Thanks from sakonpure6
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  8. #8
    Super Member sakonpure6's Avatar
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    Re: Solve for t

    alright thank you very much!
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