Are these trigonometric "statements" equal In my Calculus book I have one statement:
$$\cos(2x) = \cos^2(x)-\sin^2(x)$$
and a couple of rows down another statement is:
$$ \cos(2x) = 2 \cos^2(x) - 1 = 1 - 2 \sin^2(x).$$
Now when trying to memorize these statements i write (over and over to see if i know it by heart)
$$ \cos(2x) = \cos^2(x) - \sin^2(x) = 2 \cos^2(x)-1 = 1 - 2\sin^2(x). $$
Is this an legitimate thing to write am I misusing the equal sign?
 A: Yes, that is fine because $ \cos^2(x) + \sin^2(x) = 1 $.
A: I would recommend that you remember any one result and derive the other two from that one and the identity $ \sin^2(x)+\cos^2(x) = 1 $
I remember $ \cos^2(x) - \sin^2(x) = 1 $ and it takes very less time to get the other two and even the result in $ \tan^2(x) $.
This is the best way to memorize trigonometric results. And anyway once you start solvin questions these identities will get etched to your subconscious. 
A: You're not misusing the equal sign.
If a number $a$ equals to $b$, to $c$ and to $d$, you can always write $a=b=c=d$.
So as $\cos(2x)$ is equal to $\cos^2(x)-\sin^2(x)$, to $2\cos^2(x)-1$ and to $1 - 2\sin^2(x)$, then you can write
$$ \cos(2x) = \cos^2(x)-\sin^2(x) = 2\cos^2(x)-1 = 1 - 2\sin^2(x). $$
A: You're correct. In mathematics, the equals sign is transitive, meaning that if $a = b$ and $b = c$, then $a = c$, which is exactly what is going on here. It doesn't really change with $\sin x$ and $\cos x$ or any other trigonometric function since they are functions from $\mathbb{R}$ to $\mathbb{R}$, meaning that they return real numbers. 
