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

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- Apr 3rd 2006, 01:29 PMgarybarlow4 sin z + cos z = 0
If you could solve the equation in the title (and show me how) I'd be very greatful. Thanks.

- Apr 3rd 2006, 02:05 PMThePerfectHackerQuote:

Originally Posted by**garybarlow**

$\displaystyle 4\sin z+\cos z=0$

Thus,

$\displaystyle 4\sin z=-\cos z$

Thus,

$\displaystyle 16\sin^2 z=\cos^2 z$

Thus,

$\displaystyle 16\sin^2 z=1-\sin^2 z$

Thus,

$\displaystyle 17\sin^2 z=1$

Thus,

$\displaystyle \sin^2 z=1/17$

Thus,

$\displaystyle \sin z=\pm \sqrt{1/17}$

Now cheack,

Notice that $\displaystyle \sin z=\sqrt{1/17}$ does not check with the first equation.

However, $\displaystyle \sin z=-\sqrt{1/17}$ does.

Thus, you need to solve,

$\displaystyle \sin z=-\sqrt{1/17}$ for all $\displaystyle z$ - Apr 5th 2006, 10:01 AMmathemagic
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 - Apr 5th 2006, 01:08 PMThePerfectHackerQuote:

Originally Posted by**mathemagic**

- Apr 6th 2006, 12:08 AMmathemagic
$\displaystyle

\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 - Apr 6th 2006, 02:00 AMRebesquesa little trouble!
Well, actually when you are dealing with trigonometric complex functions, a standard trick is to substitute for exponentials. Remember

$\displaystyle

{\rm e}^{iz}-{\rm e}^{-iz}=2i{\rm sin}z, \ {\rm e}^{iz}+{\rm e}^{-iz}=2{\rm cos}z

$

and after substitution

$\displaystyle

(1-4i){\rm e}^{iz}+(1+4i){\rm e}^{-iz}=0

$ or $\displaystyle

{\rm e}^{2iz}=\frac{15-8i}{17}

$

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

so the solution is

$\displaystyle

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

$ or

$\displaystyle

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

so that

$\displaystyle 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, $\displaystyle {\rm Arg}$ the principal argument and $\displaystyle k$ an integer. - Apr 6th 2006, 02:47 AMmathemagic
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 - Apr 6th 2006, 03:01 AMRebesquesQuote:

cos(z - 1.33) = 0

- Apr 6th 2006, 03:10 AMmathemagic
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 - Apr 6th 2006, 03:50 AMRebesques
Oh, :confused: I just realised what :eek: Seems ok!

- Apr 6th 2006, 05:59 AMThePerfectHacker
I think we should keep the math simple for the high school students. Not everyone here knows advanced math.