Does $\sin^2x-\cos^2x$ equal $\cos(2x)$

Question

My professor changed $\sin^2x-\cos^2x$ to $\cos(2x)$. He often makes mistakes so I just wanted to make sure because after looking online I found out that $\sin^2x-\cos^2x$ but not sure if you can rearrange it to make to $\cos(2x)$ or would it have to be -$\cos(2x)$?

Answer 1

The identity is indeed $$\cos(2x)=\cos^2(x)-\sin^2(x)$$ and in general this is not equal to $$\sin^2(x)-\cos^2(x)=-\cos(2x)$$

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If you're familiar with De Moivre's formula, we can derive the identity as $$\cos(2x)=\Re(e^{2xi})=\Re((\cos(x)+i\sin(x))^2)=\Re(\cos^2(x)+2i\cos(x)\sin(x)-\sin^2(x))$$

Answer 2

Don't forget one deduces from the above two other relations that should also be known by heart: \begin{align} \cos 2x&=2\cos^2x-1,\\ \cos 2x&=1-2\sin^2x. \end{align}

Answer 3

Recall that

$$\cos (x+y)=\cos x \cos y - \sin x \sin y$$

to check set $y=-x$ then

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

Answer 4

No...

$$\sin^2 x - \cos^2 x \implies \sin^2 x - (1 - \sin^2 x) \implies 2 \sin^2 x - 1 \ne \cos 2x$$

however $$-\cos 2x \implies -(\cos^2 x - \sin^2 x) \implies \sin^2 x - \cos^2 x$$