Why are you allowed to do this?

[motornoob101]

So if I have something like this..

$$rcos\theta =-r^{2}sin^{2}\theta$$

I can cancel out one of the r to get

$$cos\theta = rsin^{2}\theta$$

but how come when you have something like..

$$sin^2\theta = sin\theta$$

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

$$sin\theta = 1$$

Is it because in the first example, we assume that r never = 0 so you can cancel it out where as in the $$sin\theta$$ example, it could be 0? Thanks.

 

[Hurkyl]

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.

 

[motornoob101]

Oh ok. Thanks for clearing that up! I forget that canceling is division! Silly me.

 

[symbolipoint]

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

$$\[ \begin{array}{l} r\,\cos \theta + r^2 \,\sin ^2 \theta = 0 \\ r\,\cos \theta + r^2 \,(1 - \cos ^2 \theta ) = 0 \\ r\,\cos \theta + r^2 - r^2 \,\cos ^2 \theta = 0 \\ r^2 \cos ^2 \theta - r\,\cos \theta - r^2 = 0 \\ OR \\ \cos ^2 \theta - \frac{1}{r}\cos \theta - 1 = 0 \\ \end{array} \]$$

 

[mrandersdk]

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.

 

[HallsofIvy]

"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.)