[Math] Verify identity $\sin 2x – \cot x = -\cot x \cos 2x$

Question

Verify the identity:

$\sin 2x – \cot x = – \cot x \cos 2x$

I haven't gotten very far:

$2 \sin x \cos x – \frac{\cos x}{\sin x}$ …

Answer 1

If you went backwards (which turns out to be easier),

$$\begin{align}-\cot x\cos 2x&=-\frac{\cos x}{\sin x}(1-2\sin^2x)\\ &=-\frac{\cos x}{\sin x} + 2\frac{\cos x}{\sin x}\cdot\sin^2x\\ &= -\frac{\cos x}{\sin x} + 2\cos x\sin x\\ &= -\cot x + \sin2x\end{align}$$

If you went forward,

$$\begin{align}\sin2x - \cot x &= 2\sin x \cos x - \frac{\cos x}{\sin x}\\ &= \cos x\left(2\sin x - \frac{1}{\sin x}\right)\end{align}$$ Combining the expression in the bracket using the algebraic identity $a - \frac{1}{b} = \frac{ab - 1}{b}$, this simplifies to: $$\cos x\left(\frac{2\sin^2x - 1}{\sin x}\right)$$ Using the double angle formula for $\cos$, $\cos2x = 1 - 2\sin^2x$, this becomes $$\cos x \left(\frac{-\cos 2x}{\sin x}\right)\\ = \frac{\cos x}{\sin x}(-\cos 2x)\\ = -\cot x\cos2x$$