# 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?

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For some basic information about writing math at this site see e.g. here, here, here and here. – Américo Tavares Mar 1 '13 at 14:39

Yes, that is fine because $\cos^2(x) + \sin^2(x) = 1$.

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The TeX code for cos and sin is \cos and \sin. – Américo Tavares Mar 1 '13 at 14:44
@AméricoTavares Thanks, I wasn't actually aware of that. – Noble. Mar 1 '13 at 14:46

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 $$\cos(2x) = \cos^2(x)-\sin^2(x) = 2\cos^2(x)-1 = 1 - 2\sin^2(x).$$

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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.

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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.

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