Why is $\cos 2x$ equal to $2\cos^2x-1$ and $1-2\sin^2 x$?

Question

How come the following trigonometric identity is true?

\begin{align} \cos(2x) &= \cos^2(x) - \sin^2(x) \quad\text{(I understand how they got this)} \\ &= 2\cos^2(x) - 1 \\ &= 1 - 2\sin^2(x). \end{align}

I understand how they arrived at the first line of math:

Let $a = b = x$, then

\begin{align} \cos(x+x) &= \cos(x)\cos(x) - \sin(x)\sin(x) \\ \cos(2x) &= \cos^2(x) - \sin^2(x) \end{align}

But after that I do not understand.

Thank you and have a great day,

Joseph Cummens

Answer 1

$$\cos 2x=\cos^2x-\sin^2x \tag1$$ $$\sin^2x+\cos^2x=1\tag2$$

Make use of these two formulas twice once by substituting in another expression for $\sin^2x$ and then for $\cos^2x$ into the first identity.