# Inverse Trig Function Help needed!

by chrisa88
Tags: function, inverse, trig
 P: 23 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?