Why is it that when proving trig identities, one must work both sides independently? Suppose that you have to prove the trig identity:
$$\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}=\sin\theta$$
I have always been told that I should manipulate the left and right sides of the equation separately, until I have transformed them each into something identical. So I would do:
$$\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}$$
$$=\frac{\sin\theta(1 - \sin^2\theta)}{\cos^2\theta}$$
$$=\frac{\sin\theta(\cos^2\theta)}{\cos^2\theta}$$
$$=\sin\theta$$
And then, since the left side equals the right side, I have proved the identity. My problem is: why can't I manipulate the entire equation? In this situation it probably won't make things any easier, but for certain identities, I can see ways to "prove" the identity by manipulating the entire equation, but cannot prove it by keeping both sides isolated.
I understand, of course, that I can't simply assume the identity is true. If I assume a false statement, and then derive from it a true statement, I still haven't proved the original statement. However, why can't I do this:
$$\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}\not=\sin\theta$$
$$\sin\theta - \sin^3\theta\not=(\sin\theta)(\cos^2\theta)$$
$$\sin\theta(1 - \sin^2\theta)\not=(\sin\theta)(\cos^2\theta)$$
$$(\sin\theta)(\cos^2\theta)\not=(\sin\theta)(\cos^2\theta)$$
Since the last statement is obviously false, is this not a proof by contradiction that the first statement is false, and thus the identity is true?
Or, why can't I take the identity equation, manipulate it, arrive at $(\sin\theta)(\cos^2\theta)=(\sin\theta)(\cos^2\theta)$, and then work backwards to arrive at the trig identity. Now, I start with a statement which is obviously true, and derive another statement (the identity) which must also be true - isn't that correct?
Another argument that I have heard for keeping the two sides isolated is that manipulating an equation allows you to do things that are not always valid in every case. But the same is true when manipulating just one side of the equation. In my first proof, the step
$$\frac{\sin\theta(\cos^2\theta)}{\cos^2\theta}$$
$$=\sin\theta$$
is not valid when theta is $\pi/2$, for example, because then it constitutes division by zero.
 A: 
Why can't I manipulate the entire equation?

You can. The analytical method for proving an identity consists of starting with the
identity you want to prove, in the present case
$$
\begin{equation}
\frac{\sin \theta -\sin ^{3}\theta }{\cos ^{2}\theta }=\sin \theta,\qquad \cos \theta \neq 0
\tag{1}
\end{equation}
$$
and establish a sequence of identities so that each one is a consequence of
the next one. For the identity $(1)$ to be true is enough that the following
holds
$$
\begin{equation}
\sin \theta -\sin ^{3}\theta =\sin \theta \cos ^{2}\theta   \tag{2}
\end{equation}
$$
or this equivalent one
$$
\begin{equation}
\sin \theta \left( 1-\sin ^{2}\theta \right) =\sin \theta \cos ^{2}\theta  
\tag{3}
\end{equation}
$$
or finally this last one 
$$
\begin{equation}
\sin \theta \cos ^{2}\theta =\sin \theta \cos ^{2}\theta   \tag{4}
\end{equation}
$$
Since $(4)$ is true so is $(1)$.
The book indicated below illustrates this method with the following identity
$$
\frac{1+\sin a}{\cos a}=\frac{\cos a}{1-\sin a}\qquad a\neq (2k+1)\frac{\pi 
}{2}
$$
It is enough that the following holds
$$
(1+\sin a)(1-\sin a)=\cos a\cos a
$$
or
$$
1-\sin ^{2}a=\cos ^{2}a,
$$
which is true if 
$$
1=\cos ^{2}a+\sin ^{2}a
$$
is true. Since this was proven to be true, all the previous indentities
hold, and so does the first identity.


Reference: J. Calado, Compêndio de Trigonometria, Empresa Literária
Fluminense, Lisbon, pp. 90-91, 1967.
A: You've got a pretty good handle on the situation.  It's not so much that you can't manipulate the potential identity as an equation as that, in general, most people shouldn't manipulate the potential identity as an equation.  The key part is what you said—use the manipulation to arrive at a true statement (that's your scratch-work), then work backwards to write your proof: starting with a true statement and arriving at the identity.
In your last example, since $\cos\theta$ is in the denominator, $\theta=\frac{\pi}{2}$ would not be in the domain of the identity, so it's okay to simplify to $\sin\theta$.
A: Prove the trig identity "LHS = RHS"
Given: LHS
Goal: RHS (or vise versa)
The reason that it is not valid to work on both sides at the same time (cross-multiplying, etc) is that you are not given "LHS = RHS", so there is no equation until after you have proven the trig identity.
Is it valid to use the equation "LHS = RHS" to prove the trig identity "LHS = RHS"?
Steve
A: I agree completely with Noah Stein. I just want to clarify/add the following:
Suppose the identity you are trying to show is  $L(\theta)=R(\theta)$ and it is undefined at  $\theta\in\left\{k\pi\,:\,k\in\Bbb Z\right\}$ because there is $\sin(\theta)$ in the denominator of $L(\theta)$ or $R(\theta)$ or both.
Then you can manipulate it as an equation by multiplying both sides by $\sin(\theta)$.
Suppose after several other manipulations/transformation that required no further multiplication by a variable, you arrived at $\cos(\theta)=\cos(\theta)$. Then the identity is proved for all $\theta\ne k\pi$ which is the largest possible set on which $L(\theta)$ and $R(\theta)$ are defined. Hence you proved it is an identity.
A: According to the definition "identity is a relation which is true for all values of x", so when we manipulate the trigonometric identity just like the trigonometric equations try to find the angles and we ended up a true relation like 0=0 independent of angle means the relation is an identity.
A: I think it can be done.  We are proving that LHS =RHS,  as by assuming so,  you arrive at a Universal truth. 
Similarly,  by assuming it's not true,  we arrive at a contradiction ,  which is called,  proof by contradiction. 
