Three Forms of cos 2x

With B = A, $\cos\left(A+B\right) = \cos\left(A\right)\cos\left(B\right)-\sin\left(A\right)\sin\left(B\right)$ gives $\cos\left(2x\right) = {\cos\left(x\right)}^{2}-{\sin\left(x\right)}^{2}$. Replace one square using sin²x + cos²x = 1 and two more forms appear:

$$\cos\left(2x\right) = 2{\cos\left(x\right)}^{2}-1$$

$$\cos\left(2x\right) = 1-2{\sin\left(x\right)}^{2}$$

All three are equal; choose the one that cancels something. In $1+\cos\left(2x\right)$ the form with -1 is the one to use: $1+2{\cos\left(x\right)}^{2}-1$, the ones cancel, and $2{\cos\left(x\right)}^{2}$ is left. In $1-\cos\left(2x\right)$ it is the form that starts with 1 that cancels.