Proofs of Trigonometric Identities II, cos 2x = 2cos^2 x - 1 = 1 - 2sin^2 x = cos^2 x - sin^2 x

Statement 1: $$\cos 2x = \cos^2 x - \sin^2 x$$

Proof 1: Use the Angle Addition Formula for Cosine:

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

Statement 2: $$\cos 2x = 1 - 2\sin^2 x$$

Proof 2: We can prove that

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

because the left-hand side is equivalent to $$\cos(2x)$$. Add $$2\sin^2(x)$$ to both sides of the equation:

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

This is obviously true.

Statement 3: $$\cos 2x = 2\cos^2 x - 1$$

Proof: It suffices to prove that

$$1 - 2\sin^2 x = 2\cos^2 x - 1$$

Add $$1$$ to both sides of the equation:

$$2 - 2\sin^2 x = 2\cos^2 x$$

Now add $$2\sin^2 x$$ to both sides of the equation:

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

Therefore this equality also holds.