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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?
Yes, that is fine because $ \cos^2(x) + \sin^2(x) = 1 $.
\cos and \sin.
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– Américo Tavares
Mar 1 '13 at 14:44
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 $$ \cos(2x) = \cos^2(x)-\sin^2(x) = 2\cos^2(x)-1 = 1 - 2\sin^2(x). $$
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.
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.
