Simplifying Trig Equation for B w/ Alpha & Beta

[neutrino2063]

I need to somehow simplify:

\frac{1}{B^2}=\frac{1+\cos{2\alpha}}{2k_1}+\frac{\sin{2\alpha}}{2k_2}+\frac{\alpha}{k_2}

to:

B=\sqrt{\frac{2}{L}}\sqrt{\frac{\beta}{1+\beta}}

Where:

\alpha=\frac{L}{2}k_2 and \beta=\frac{L}{2}k_1

And \beta is also defined transcendentally:

\beta=\alpha\tan{\alpha}

Any ideas would be appreciated, I see no way of getting rid of the trig functions. I've tried looking for identities and even given it to mathematica; it seems to me I'm missing some sort of special trick.

 

[Mark44]

Is it a typo that alpha and beta are equal? It seems an unnecessary complication to add another variable if it's not needed. Otherwise I would just start substituting things into the right side of your first equation and see where that takes me.

 

[neutrino2063]

Ah, it is... thanks, it's fixed now alpha should be (L/2)*k2

 

[Mark44]

Now replaces cos(2 alpha) and sin(2 alpha) using the double-angle identities, and use your other two relationships to get rid of alpha to see if you can make the right side look like the left.