Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why are you allowed to do this?

  1. Apr 1, 2008 #1
    So if I have something like this..

    [tex]rcos\theta =-r^{2}sin^{2}\theta[/tex]

    I can cancel out one of the r to get

    [tex]cos\theta = rsin^{2}\theta[/tex]

    but how come when you have something like..

    [tex]sin^2\theta = sin\theta[/tex]

    and say you are trying to find the zeros of this equation, you can't just do

    [tex]sin\theta = 1[/tex]

    Is it because in the first example, we assume that r never = 0 so you can cancel it out where as in the [tex]sin\theta [/tex] example, it could be 0? Thanks.
  2. jcsd
  3. Apr 1, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You are right -- you can never divide by an expression that may be zero, and cancelling is a form of division.

    Incidentally, I would have said that both of those examples of cancelling are illegal. You can only do the first one if r is nonzero, but that is not generally true! It is, of course, legal whenever you do happen to know that r is nonzero -- for example, if you happen to split a problem into two cases, one where r is zero, and one where r is nonzero, then clearly in the second case, you'd be allowed to cancel an r.
  4. Apr 1, 2008 #3
    Oh ok. Thanks for clearing that up! I forget that canceling is division! Silly me.
  5. Apr 2, 2008 #4


    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    Thanks for the fun example, motonoob101. You will have a quadratic equation with variable of cosine of theta:

    [tex] \[
    r\,\cos \theta + r^2 \,\sin ^2 \theta = 0 \\
    r\,\cos \theta + r^2 \,(1 - \cos ^2 \theta ) = 0 \\
    r\,\cos \theta + r^2 - r^2 \,\cos ^2 \theta = 0 \\
    r^2 \cos ^2 \theta - r\,\cos \theta - r^2 = 0 \\
    OR \\
    \cos ^2 \theta - \frac{1}{r}\cos \theta - 1 = 0 \\
  6. Apr 3, 2008 #5
    also note that in you first example you "loose" a solution when deviding with r, namely r = 0, just like you loose solutions when deviding by sin in the second example.
  7. Apr 3, 2008 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    "lose", not "loose".

    (I don't know why that irks me so much more than other misspellings! Perhaps because "loose" is a perfectly good word, just the wrong one.)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?