I see the difference! I'm a little new here for thanks for the tip.
I believe I was able to solve the problem with your help. It is very much appreciated! Thanks for the expedient responses and have a nice holiday season.
I did differentiate the LHS wrt θ and the RHS wrt phi. Come to think of it, that doesn't make sense at all!
If I were to solve for \theta, I think I'd get:
\theta=2arcsin(ksin(\phi))
then differentiating with respect to phi,
\frac{d\theta}{d\phi}=2\frac{kcos(\phi)}{\sqrt{1-k^2sin(\phi)}}
so...
Differenting both sides and then doing some algebra gives me
d\theta=\frac{2sin(\alpha/2)cos(\phi)d\phi}{cos(\theta/2)}
Also thank you for pointing out that my denominator can be rewritten as kcos(θ). That seems potentially helpful.
If sin(\theta/2)=ksin(\phi), then...
Homework Statement
given: dt=-\frac{1}{2}\sqrt{\frac{l}{g}}\frac{d\theta}{\sqrt{sin^2(\alpha/2)-sin^2(\theta/2)}}
make the change of variables sin(\theta/2)=sin(\alpha/2)sin(\phi)
to show that: dt=-\sqrt{\frac{l}{g}}\frac{d\phi}{\sqrt{1-k^2sin^2(\phi)}}
where k=sin(\alpha/2)
Homework...