4 sin z + cos z = 0

**garybarlow** · April 3rd 2006, 01:29 PM

If you could solve the equation in the title (and show me how) I'd be very greatful. Thanks.

**ThePerfectHacker** · April 3rd 2006, 02:05 PM

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 $z$

**mathemagic** · April 5th 2006, 10:01 AM

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

**ThePerfectHacker** · April 5th 2006, 01:08 PM

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.

**mathemagic** · April 6th 2006, 12:08 AM

$ \cos z\not=0 $
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

**Rebesques** · April 6th 2006, 02:00 AM

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

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

(check this please! calculations are not my strong point

)

so the solution is

$ 2iz={\rm log}\left(\frac{15-8i}{17}\right) $ or

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

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

where the logarithm is the usual real one, ${\rm Arg}$ the principal argument and $k$ an integer.

**mathemagic** · April 6th 2006, 02:47 AM

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

**Rebesques** · April 6th 2006, 03:01 AM

cos(z - 1.33) = 0

I have no idea how this can be solved without exponentials.

**mathemagic** · April 6th 2006, 03:10 AM

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

**Rebesques** · April 6th 2006, 03:50 AM

Oh,

I just realised what

Seems ok!

**ThePerfectHacker** · April 6th 2006, 05:59 AM

I think we should keep the math simple for the high school students. Not everyone here knows advanced math.

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