Reasoning that $ \sin2x=2 \sin x \cos x$ In mathcounts teacher told us to use the formula $ \sin2x=2 \sin x \cos x$.
What's the math behind this formula that made it true?  Can someone explain?

A rotation matrix is clearly a linear transformation, so by rotating the elementary basis vectors, a two-dimensional rotation matrix for angle $x$ is
\begin{align}
\begin{bmatrix}
\cos(x) & -\sin(x) \\
\sin(x) & \cos(x)
\end{bmatrix}.
\end{align}
Squaring this matrix is equivalent to applying the rotation twice, so
\begin{align}
\begin{bmatrix}
\cos(2x) & -\sin(2x) \\
\sin(2x) & \cos(2x)
\end{bmatrix}
&=
\begin{bmatrix}
\cos^2(x) - \sin^2(x) & -2\sin(x)\cos(x) \\
2\sin(x)\cos(x) & \cos^2(x) - \sin^2(x)
\end{bmatrix},
\end{align}
which implies that
\begin{align}
\cos(2x) &= \cos^2(x) - \sin^2(x) \\
\sin(2x) &= 2\sin(x)\cos(x).
\end{align}
Using rotation matrices with different angles gives some of the double-angle (or triple-angle, etc.) identities.
