Rearranging Trigonometric Functions

  • #1
This is my very first post - so i hope I don't break any rules - its more of a formula rearranging question/confirmation so here goes...

Homework Statement


Currently working on friction - static/kinetic - so in my text book it states in a side bar "info bit" that

tanθ=sinθ/cosθ my question is : can this equation be rearranged - see below...


Homework Equations



so if tanθ=sinθ/cosθ then does sinθ=(tanθ)(cosθ) and does cosθ=sinθ/tanθ


The Attempt at a Solution



I haven't come across any questions so far where i would need to use sinθ=(tanθ)(cosθ) and cosθ=sinθ/tanθ - more or less a general question of would I encounter having to use the sinθ=(tanθ)(cosθ) and cosθ=sinθ/tanθ rather than tanθ=sinθ/cosθ - numerically it works, but are these variations used?


Hope this wasn't clear as mud... If i can provide any more info - let me know!

:bugeye:
 

Answers and Replies

  • #2
collinsmark
Homework Helper
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Hello talknerdy2me,

Welcome to Physics Forums! :smile:

This is my very first post - so i hope I don't break any rules - its more of a formula rearranging question/confirmation so here goes...

Homework Statement


Currently working on friction - static/kinetic - so in my text book it states in a side bar "info bit" that

tanθ=sinθ/cosθ my question is : can this equation be rearranged - see below...
Yes.

Homework Equations



so if tanθ=sinθ/cosθ then does sinθ=(tanθ)(cosθ) and does cosθ=sinθ/tanθ
Yep. You got it. :smile:

The Attempt at a Solution



I haven't come across any questions so far where i would need to use sinθ=(tanθ)(cosθ) and cosθ=sinθ/tanθ - more or less a general question of would I encounter having to use the sinθ=(tanθ)(cosθ) and cosθ=sinθ/tanθ rather than tanθ=sinθ/cosθ - numerically it works, but are these variations used?


Hope this wasn't clear as mud... If i can provide any more info - let me know!

:bugeye:
If it helps, this might make it a little more easy to remember:

[tex] \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} [/tex]
[tex] \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} [/tex]
[tex] \tan \theta = \frac{\text{opposite}}{\text{adjacent}} [/tex]
If you can remember those, the other relationships you mentioned above should fall into place.
 
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