\(\sin 2x = 2\sin x\cos x\quad\cos 2x = \cos^2\!x - \sin^2\!x\)
Pythagorean: \(\sin^2\theta+\cos^2\theta=1\)
📌 For equations, factorise — don't divide both sides by a trig function (you lose solutions).
Example
Solve \(\sin 2x = \sin x\) for \(x \in [0, 2\pi)\)
\(2\sin x\cos x - \sin x = 0 \Rightarrow \sin x(2\cos x - 1)=0\)
\(x = 0,\; \tfrac{\pi}{3},\; \pi,\; \tfrac{5\pi}{3}\)