Just a few notes on how to proceed on the penultimate homework of the semester.

We're to show that the solutions for the 30/60/90 triangular waveguide given in the last homework set will also work for a waveguide that's formed from an equilateral traingle. The three corners of the equilateral traingle are located at $\left(x,y\right) = \left(0, 0\right)$, $\left(x,y\right) = \left(a, a/\sqrt{3}\right)$, and $\left(x,y\right) = \left(a, -a/\sqrt{3}\right)$.

This falls out immediately from last week's homeowrk. Because the sine function is peiodic in $\pi$ over the domain from $\left(-\infty, \infty\right)$, the solution given last week in terms of sines will still evaluate to zero on the wall that falls at negative $y$. coordites. The positive $x$ coordinates of the functions will evaluate to 0 on the wall in the same manner they did before??? There's an issue here. It's products of the $x$ and $y$ sinusoids that all sum to zero. These will need to be checked to determine if they still evaluate to zero. Since the y terms are converted to x terms using the coordinates of the walls, there's very little to do in checking that things are stil OK. Everyting converts down to a difference of cosines expressed as

$cos\dfrac{m-n}{3a} - cos\dfrac{n-m}{3a}$

However, because the three $y$ sin functions are odd, they will merely wind up with negative signs out in front that can be factored out of the entire expression. The result will be the same.

We're to show that the solutions for the 30/60/90 triangular waveguide given in the last homework set will also work for a waveguide that's formed from an equilateral traingle. The three corners of the equilateral traingle are located at $\left(x,y\right) = \left(0, 0\right)$, $\left(x,y\right) = \left(a, a/\sqrt{3}\right)$, and $\left(x,y\right) = \left(a, -a/\sqrt{3}\right)$.

This falls out immediately from last week's homeowrk. Because the sine function is peiodic in $\pi$ over the domain from $\left(-\infty, \infty\right)$, the solution given last week in terms of sines will still evaluate to zero on the wall that falls at negative $y$. coordites. The positive $x$ coordinates of the functions will evaluate to 0 on the wall in the same manner they did before??? There's an issue here. It's products of the $x$ and $y$ sinusoids that all sum to zero. These will need to be checked to determine if they still evaluate to zero. Since the y terms are converted to x terms using the coordinates of the walls, there's very little to do in checking that things are stil OK. Everyting converts down to a difference of cosines expressed as

$cos\dfrac{m-n}{3a} - cos\dfrac{n-m}{3a}$

However, because the three $y$ sin functions are odd, they will merely wind up with negative signs out in front that can be factored out of the entire expression. The result will be the same.

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