Use Square Rule: \({i}^{2}=-1\).
\[\tan{(varph\times -1)}+6=6.045\]
Simplify \(varph\times -1\) to \(-varph\).
\[\tan{(-varph)}+6=6.045\]
Subtract \(6\) from both sides.
\[\tan{(-varph)}=6.045-6\]
Simplify \(6.045-6\) to \(0.045\).
\[\tan{(-varph)}=0.045\]
Ask: What values of \(-varph\) will make \(\tan{(-varph)}\) equal \(0.045\)?
\[-varph=\tan^{-1}{(0.045)},\tan^{-1}{(0.045)}+\pi \]
Since tan is a periodic function, add the periodicity.
\[\begin{aligned}&-varph=\pi n+\tan^{-1}{(0.045)},n \in Z\\&-varph=\pi n+\tan^{-1}{(0.045)}+\pi ,n \in Z\end{aligned}\]
The above equations can be expressed as a single equation.
\[-varph=\pi n+\tan^{-1}{(0.045)},n \in Z\]
Solve for \(v\).
\[v=-\frac{\pi n+\tan^{-1}{(0.045)}}{arph},n \in Z\]
v=sequence(-(PI*n+arctan(0.045))/(a*r*p*h),in(n,Z))
