Rearranging Trigonometric Functions

AI Thread Summary
The discussion focuses on the rearrangement of the trigonometric identity tanθ = sinθ/cosθ. It confirms that sinθ can be expressed as (tanθ)(cosθ) and cosθ as sinθ/tanθ, validating these transformations. The original poster inquires about the practical use of these rearrangements in problems, noting they haven't encountered them yet. The response emphasizes that understanding the basic definitions of sine, cosine, and tangent can aid in remembering these relationships. Overall, the variations are mathematically correct and can be useful in certain contexts.
talknerdy2me
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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 textbook 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:
 
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Hello talknerdy2me,

Welcome to Physics Forums! :smile:

talknerdy2me said:
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 textbook 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:

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