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Math Help - how do you solve this?

  1. #1
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    how do you solve this?

    6. 5(sin 6t - 4cos 4t) = 0

    15. sin2x = cosx

    (on the interval [0, 2pi])

    Thanks!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by finalfantasy View Post
    6. 5(sin 6t - 4cos 4t) = 0
    we have \sin 6t - 4 \cos 4t = 0

    use the formulas

    \sin nt = \sin [(n - 1)t + t] = \sin (n - 1)t \cos t + \sin t \cos (n - 1)t

    and

    \cos nt = \cos [(n - 1)t + t] = \cos (n - 1)t \cos t - \sin (n - 1)t \sin t

    to keep breaking down the sines and cosines until you get them to \sin 2t and \cos 2t. then use the double angle formulas. then simplify and continue. it will be a lot of work, but it is the kind of work where you can keep following your nose


    15. sin2x = cosx

    (on the interval [0, 2pi])
    use the double angle formulas for sine

    \sin 2x = \cos x

    \Rightarrow 2 \sin x \cos x = cos x

    \Rightarrow 2 \sin x \cos x - \cos x = 0

    \Rightarrow \cos x ( 2 \sin x - 1) = 0

    now continue


    EDIT: This is my 53th post!!!!!!!
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by finalfantasy View Post
    6. 5(sin 6t - 4cos 4t) = 0
    Start with dividing both sides by 5 so you have
    sin(6t) - 4sin(4t) = 0

    I'd simplify this a little bit and sub in y = 2x:
    sin(3y) - 4sin(2y) = 0

    Now,
    sin(2y) = 2sin(y)cos(y)

    sin(3y) = sin(2y + y) = sin(2y)cos(y) + sin(y)cos(2y)

    = 2sin(y)cos^2(y) + sin(y) - 2sin^3(y)

    = 2sin(y)(1 - sin^2(y)) + sin(y) - 2sin^3(y)

    = 3sin(y) - 4sin^3(y)

    So we need to solve
    (3sin(y) - 4sin^3(y)) - 4(2sin(y)cos(y)) = 0

    -4sin^3(y) + 3sin(y) - 8sin(y)cos(y) = 0

    You can immediately factor out a sin(y), so y = 0, \pi \implies x = 0, \frac{\p}{2} are solutions.

    Now we need to solve
    -4sin^2(y) + 3 - 8cos(y) = 0

    Well,
    -4(1 - cos^2(y)) + 3 - 8cos(y) = 0

    -4 - 4cos^2(y) + 3 - 8cos(y) = 0

    4cos^2(y) +  8cos(y) +1 = 0
    which is a quadratic in cos(y). Can you take over?

    -Dan
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