sin 2x = 2 sin x cos x

Put B = A into the sum formula: $\sin\left(A+A\right) = \sin\left(A\right)\cos\left(A\right)+\cos\left(A\right)\sin\left(A\right)$. The two products are the same, so

$$\sin\left(2x\right) = 2\sin\left(x\right)\cos\left(x\right)$$

Doubling the angle does not double the sine: at 30°, $\sin\left(60^{\circ}\right)$ is about 0.87, not $2\sin\left(30^{\circ}\right) = 1$. The 2 comes out only together with the cos x.

Take $\frac{\sin\left(2x\right)}{\cos\left(x\right)}$. Write the top as $2\sin\left(x\right)\cos\left(x\right)$; the bottom cos x is now a factor of the top, so it cancels and leaves $2\sin\left(x\right)$, wherever cos x is not 0.