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Math Help - 4 sin z + cos z = 0

  1. #1
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    4 sin z + cos z = 0

    If you could solve the equation in the title (and show me how) I'd be very greatful. Thanks.
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  2. #2
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    Quote Originally Posted by garybarlow
    If you could solve the equation in the title (and show me how) I'd be very greatful. Thanks.
    You have,
    4\sin z+\cos z=0
    Thus,
    4\sin z=-\cos z
    Thus,
    16\sin^2 z=\cos^2 z
    Thus,
    16\sin^2 z=1-\sin^2 z
    Thus,
    17\sin^2 z=1
    Thus,
    \sin^2 z=1/17
    Thus,
    \sin z=\pm \sqrt{1/17}
    Now cheack,
    Notice that \sin z=\sqrt{1/17} does not check with the first equation.
    However, \sin z=-\sqrt{1/17} does.
    Thus, you need to solve,
    \sin z=-\sqrt{1/17} for all z
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  3. #3
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    4 sin z + cos z = 0

    would it not be easier to divide by cos z to get

    4 tan z + 1 = 0
    and so you'd have to solve tan z = -1/4
    that seems easier
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  4. #4
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    Quote Originally Posted by mathemagic
    4 sin z + cos z = 0

    would it not be easier to divide by cos z to get

    4 tan z + 1 = 0
    and so you'd have to solve tan z = -1/4
    that seems easier
    You first need to show that, \cos z\not=0 (which is true), yes it is of course easier.
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  5. #5
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    <br />
\cos z\not=0<br />
    because whenever cos z is 0, 4 sin z is either 4 or -4, and so the expression would be -4 + 0 or 4 + 0, neither of which are 0, so you can divide by cos z
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  6. #6
    Super Member Rebesques's Avatar
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    a little trouble!

    Well, actually when you are dealing with trigonometric complex functions, a standard trick is to substitute for exponentials. Remember
    <br />
 {\rm e}^{iz}-{\rm e}^{-iz}=2i{\rm sin}z, \ {\rm e}^{iz}+{\rm e}^{-iz}=2{\rm cos}z<br />

    and after substitution
    <br />
 (1-4i){\rm e}^{iz}+(1+4i){\rm e}^{-iz}=0<br />
or <br />
 {\rm e}^{2iz}=\frac{15-8i}{17}<br />

    (check this please! calculations are not my strong point )

    so the solution is

    <br />
 2iz={\rm log}\left(\frac{15-8i}{17}\right)<br />
or

    <br />
 z=-\frac{i}{2}{\rm log}\left(\frac{15-8i}{17}\right)
    so that

    z=<br />
-\frac{i}{2}<br />
\left({\rm log}\left|\frac{15-8i}{17}\right|+i{\rm Arg}\left(\frac{15-8i}{17}\right)+2k\pi i\right),

    where the logarithm is the usual real one, {\rm Arg} the principal argument and k an integer.
    Last edited by Rebesques; April 6th 2006 at 03:05 AM.
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  7. #7
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    I've not learnt about the exponential stuff you can do with trigonometric equations like that yet.
    I thought of another way to solve it

    4 sin z + cos z = 0

    let:
    4 sin z + cos z = k cos(z - a)

    4 sin z + cos z = k cos a cos z + k sin a sin z

    k sin a = 4
    k cos a = 1
    => tan a = 4
    a = 1.33

    4 sin z + cos z = k cos(z - 1.33)

    k = sqrt(4^2 + 1^2)
    k = sqrt(17)

    4 sin z + cos z = sqrt(5) cos(z - 1.33)
    so now you need to solve
    sqrt(17) cos(z - 1.33) = 0
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  8. #8
    Super Member Rebesques's Avatar
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    cos(z - 1.33) = 0
    I have no idea how this can be solved without exponentials.
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  9. #9
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    to solve sqrt(17)cos (z - 1.33) = 0
    you do this:

    z- 1.33 = cos -1 (0)
    = pi/2, 3pi/2 ...
    z = pi/2 + 1.33, 3pi/2 + 1.33 etc
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  10. #10
    Super Member Rebesques's Avatar
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    Oh, I just realised what Seems ok!
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  11. #11
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    I think we should keep the math simple for the high school students. Not everyone here knows advanced math.
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