# Inverse Trig Function Help needed!

1. ### chrisa88

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?

2. ### UltrafastPED

1,916
Your math expressions are yielding an error here.

Just as there are many trigonometric relationships, so there are apparently just as many relations between their inverses. See http://en.wikipedia.org/wiki/Inverse_trigonometric_functions

As always, start with the easiest, defining relationships and build out from there. Note that as sine and cosine are related by complementary angles, so are their inverses.

3. ### chrisa88

23
I thought this was an error, but the solutions manual to my quantum mechanics class AND the handwritten solutions provided by my professor both have this error. Thank you for confirming!!