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## Main Question or Discussion Point

How does this work? I'm very confused about the phi is solved using inverse sin.

knowing: A=(c[itex]^{2}_{1}[/itex]+c[itex]^{2}_{2}[/itex])[itex]^{1/2}[/itex] and c[itex]_{2}[/itex]= Acos([itex]\phi[/itex])

solve for [itex]\phi[/itex]

which yields: [itex]\phi[/itex]=sin[itex]^{-1}[/itex][itex]\frac{c_{2}}{(c^{2}_{1}+c^{2}_{2})^{1/2}}[/itex]=tan[itex]^{-1}[/itex][itex]\frac{c_{2}}{c_{1}}[/itex]

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[itex]^{-1}[/itex]. Where am I going wrong?

knowing: A=(c[itex]^{2}_{1}[/itex]+c[itex]^{2}_{2}[/itex])[itex]^{1/2}[/itex] and c[itex]_{2}[/itex]= Acos([itex]\phi[/itex])

solve for [itex]\phi[/itex]

which yields: [itex]\phi[/itex]=sin[itex]^{-1}[/itex][itex]\frac{c_{2}}{(c^{2}_{1}+c^{2}_{2})^{1/2}}[/itex]=tan[itex]^{-1}[/itex][itex]\frac{c_{2}}{c_{1}}[/itex]

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[itex]^{-1}[/itex]. Where am I going wrong?