Plane Wave Propogation (help understanding steps in book)

[FrogPad]

Plane Wave Propagation (help understanding steps in book)

Basically this problem requires that \vec H vanishes at t = 3ms. I don't understand how they did the last step, here is the solution:

\vec H = \hat z 4 \times 10^{-6} \cos \left(10^7 \pi t - k_0 y +\frac{\pi}{4} \right)

k_0 = \omega \sqrt{\mu_0 \epsilon_0} = \frac{10^7 \pi}{3 \times 10^8} = 0.105 \,\,\,\, (rad/m)
\lambda = \frac{2\pi}{k_0} = 60 \,\,\,\, (m)

At t = 3 \times 10^{-3} \,\,\,\, (s) we require the argument of cosine in \vec H:

10^7 \pi (3 \times 10^{-3}) - \frac{\pi}{30}y +\frac{\pi}{4} = \pm n \pi + \frac{\pi}{2}, \,\,\,\,n=0,1,2,\ldots

y = \pm 30n - 7.5 \,\,\,\, (m)

Now how the hell do they get the last line? The... y = \pm 30n - 7.5 \,\,\,\, (m) (Lets call this (1)) line.
They drop 10^7 \pi (3 \times 10^{-3}) why?

My problem is that if we plug in (1) into,
\vec H = \hat z 4 \times 10^{-6} \cos \left(10^7 \pi t - k_0 y +\frac{\pi}{4} \right)

and set t = 0.003, then if we feed the function values of n then the cosine doesn't drop to 0.

Thanks in advance!

 

[OlderDan]

The term they dropped is a multiple of pi (and in fact a multiple of 2pi). Adding a multiple of pi to an angle at which the cosine is zero is still going to give you a cosine of zero.

 

[FrogPad]

The term they dropped is a multiple of pi (and in fact a multiple of 2pi). Adding a multiple of pi to an angle at which the cosine is zero is still going to give you a cosine of zero.

Ahh... tricky tricky. (I mean, yeah, it's not too tricky but still, I guess I will add this to my list of things to look for subconsciously while evaluating expressions)