# Trying to prove a trigonometric identity

I've been trying to solve it for quite some time but I still don't get it why it is true.

The original equation is: \begin{equation*} 1-\frac{\sin{^2}\theta}{1-\cos\theta}=-\cos\theta. \end{equation*} My work so far: \begin{equation*} \frac{1-\cos\theta}{1-\cos\theta}-\frac{\sin{^2}\theta}{1-\cos\theta} =\frac{1-\cos\theta-\sin{^2}\theta}{1-\cos\theta} =\frac{(1-\sin{^2}\theta)-\cos\theta}{1-\cos\theta} \\ =\frac{\cos{^2}\theta-\cos\theta}{1-\cos\theta} =\frac{\cos\theta(\cos\theta-1)}{1-\cos\theta}. \end{equation*} I saw on some sites that this is equal to $-\cos\theta$ but I don't see why.

• You've got $\cos(\theta)-1$ in the numerator which is $-1\cdot(1-\cos(\theta))$. Now cancel and you're done. – Michael Burr May 14 '15 at 15:54
• $\cos x-1=-(1-cos x)$ – E.H.E May 14 '15 at 15:54

$\textbf{hint}$ $$\cos\theta -1 = -(1-\cos\theta)$$

• Got it. Thanks! – Dormin May 14 '15 at 17:20

You're very close. From where you left off

$\frac{\cos\theta(\cos\theta-1)}{1-\cos\theta}$

=$\frac{\cos\theta(-1)(1-\cos\theta)}{1-\cos\theta}$

=${\cos\theta(-1)}$

=${-\cos\theta}$

• Some equals signs would be nice. – N. F. Taussig May 15 '15 at 20:30

Shorter: $$1-\frac{\sin{^2}\theta}{1-\cos\theta}=-\cos\theta\iff 1+\cos\theta=\frac{\sin{^2}\theta}{1-\cos\theta}\iff 1-\cos^2\theta=\sin^2\theta.$$