Recent content by llama123

Wonderful! Many thanks! It works perfectly, all I needed to use was: \cos^3(\phi) = (3\cos(\phi)+\cos(3\phi))/4 to finish the job. Thanks again.

Thanks for your reply. Yes you are right, excuse me for the error, it was a mistake in copying it from my notes. So now the problem is how to go from this: \cos(\psi) = \frac{1}{\sqrt{1+(\frac{q + \delta q}{q} \tan(\hat{\psi}))^2}} To the result: \cos(\psi) = (1-\frac{\delta q}{4q})...

Sorry for the mistake, now it should be OK. Hi all, I am reading a paper related to astrophysics and I am stuck in a step in a calculation. It is about orbiting gas in a spiral galaxy and it calculates the errors in the fitted rotation velocity if one of the viewing angles is incorrectly...

Hi all, I am reading a paper related to astrophysics and I am stuck in a step in a calculation. It is about orbiting gas in a spiral galaxy and it calculates the errors in the fitted rotation velocity if one of the viewing angles is incorrectly estimated. My problem is that I don't understand...