Inverse Trig Function Help needed!

  • Thread starter chrisa88
  • Start date
  • #1
chrisa88
23
0
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
Gold Member
1,912
216
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
0
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!!
 

Suggested for: Inverse Trig Function Help needed!

  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
16
Views
726
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
2K
Replies
3
Views
2K
Replies
3
Views
7K
Replies
7
Views
5K
  • Last Post
Replies
5
Views
6K
Replies
7
Views
8K
Top