# Inverse Trig Function Help needed!

## Main Question or Discussion Point

How does this work? I'm very confused about the phi is solved using inverse sin.
knowing: A=(c$^{2}_{1}$+c$^{2}_{2}$)$^{1/2}$ and c$_{2}$= Acos($\phi$)
solve for $\phi$
which yields: $\phi$=sin$^{-1}$$\frac{c_{2}}{(c^{2}_{1}+c^{2}_{2})^{1/2}}$=tan$^{-1}$$\frac{c_{2}}{c_{1}}$
I'm not sure how we use the inverse sin to find the phi in the cos function.
I thought to get the inside of the parenthesis of cos you would use inverse cos, or cos$^{-1}$. Where am I going wrong?

UltrafastPED