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mmwave
Jan10-04, 06:47 PM
After a little calculus and lots of algebra I get the expression

cot(2la) = (l^2 - k^2) / (2kl)

it is suggested to simplify with the trig identity

tan(t/2) = sqrt(1 + cot^2(t) ) - cot(t)

when I do so, I get

tan(la) = k/l

notice that we have gone from cot(2la) to tan(la). The text I am reading says

tan(la) = l/k

which I think is wrong. Can someone else please give it a try?

the key step was combining the 1 + cot^2(t) over a common denom.

himanshu121
Jan11-04, 06:09 AM
i believe u might have missed out at other roots I get both -l/k as well as k/l

cot2la=\frac{1-tan^22la}{2tanla}=\frac{l^2-k^2}{2kl}

After rearrangement u will get

kltan^2la+(l^2-k^2)tanla-kl=0

so two roots will be given by

tanla=\frac{-(l^2-k^2) \pm \sqrt{(l^2-k^2)^2+4k^2l^2}}{2kl}

After solving u will get the 2 roots

tanla = \frac{k}{l} \ & \ -\frac{l}{k}

mmwave
Jan11-04, 06:51 PM
Thanks. There is a second root but on that one the book & I agree so I assumed I was correct. [:)]

Also, I like you method better than the obscure substitution suggested by the book.

himanshu121
Jan12-04, 09:51 AM
[:))]