Re: A problem with cotangent squared
Hello, Vix!
Since we don't have a \(\displaystyle \fbox{\cot}\) key on our calculator,
. . I would solve for \(\displaystyle \theta\) like this . . .
We have: \(\displaystyle \L\:\frac{5}{3} \;=\;\cot^2\left(\frac{\theta}{2}\right)\)
. . . . . . . . \(\displaystyle \L\frac{5}{3}\;=\;\frac{1}{\tan^2\left(\frac{\theta}{2}\right)}\)
\(\displaystyle \L\frac{5}{3}\cdot\tan^2\left(\frac{\theta}{2}\right) \;=\;1\)
.. \(\displaystyle \L\tan^2\left(\frac{\theta}{2}\right) \;=\;\frac{3}{5} \;=\;0.6\)
. . \(\displaystyle \L\tan\left(\frac{\theta}{2}\right) \;=\;\pm\sqrt{0.6}\)
. . . . . . . .\(\displaystyle \L\frac{\theta}{2}\;=\;\tan^{-1}\left(\pm\sqrt{0.6}\right)\)
. . . . . . . .\(\displaystyle \L\theta \;= \;2\cdot\tan^{-1}\left(\pm\sqrt{0.6}\right)\)