# Rearranging Trigonometric Functions

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!

collinsmark
Homework Helper
Gold Member
Hello talknerdy2me,

Welcome to Physics Forums!

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.

## 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!

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.

berkeman