Inverse Trig Function Help needed!

  • Thread starter chrisa88
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  • #1
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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?
 

Answers and Replies

  • #2
UltrafastPED
Science Advisor
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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
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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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