
#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

Thanks
P: 1,294

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!!



Register to reply 
Related Discussions  
Inverse Trig Function: Find Derivative of the Function  Calculus & Beyond Homework  11  
Inverse trig function  Precalculus Mathematics Homework  8  
Inverse Trig Function  Precalculus Mathematics Homework  5  
inverse trig function  Calculus & Beyond Homework  2  
Another inverse trig function  Calculus & Beyond Homework  1 