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Inverse Trig Function Help needed! 
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#1
Sep1213, 11:55 PM

P: 23

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? 


#2
Sep1313, 01:07 AM

Sci Advisor
Thanks
PF Gold
P: 1,908

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...tric_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
Sep1313, 01:09 AM

P: 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!!



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