$$\dfrac{\sin(2x)}{1-\cos (2x)}$$
How do I simplify given expression?
My attempt:
We know the double-angle identity for $\sin(2x)$ and $\cos(2x)$ as shown below
$$\sin(2x) = 2\sin (x)\cos(x)$$
$$\cos(2x) = 2\cos(x)-1$$
So we have that
$$\dfrac{2\sin (x)\cos(x)}{1-2\cos(x)-1} = \dfrac{2\sin(x)\cos(x)}{-2\cos^2(x)} = \dfrac{2\sin(x)}{-2\cos(x)} = -\tan (x)$$
I believe I've gone wrong somewhere.
$\endgroup$ 13 Answers
$\begingroup$We have
$$\cos(2x) = \cos^2x-\sin^2x=2\cos^\color{red}2(x)-1=1-2\sin^2 x$$
and then
$$\dfrac{\sin(2x)}{1-\cos (2x)}=\dfrac{2\sin x \cos x}{1-1+2\sin^2 x}=\cot x$$
$\endgroup$ 7 $\begingroup$You got one of the signs mixed up.
$$\frac{\sin(2x)}{1-\cos (2x)} = \dfrac{\sin(2x)}{1-(2\cos^2x-1)} = \frac{\sin(2x)}{1-2\cos^2x+1}$$
Instead, you accidentally used
$$\frac{\sin(2x)}{1-\cos (2x)} \color{red}{\neq \frac{\sin(2x)}{1-2\cos^2x-1}}$$
As for the simplification itself,
$$\frac{\sin(2x)}{1-\cos (2x)} = \frac{2\sin x\cos x}{1-(1-2\sin^2x)} = \frac{2\sin x\cos x}{2\sin^2 x} = \frac{\cos x}{\sin x} = \cot x$$
Also, to point out, you used $\cos(2x) = 2\cos^2x-1$, which doesn’t actually get you anywhere since you get two $+1$’s. Using $\cos^2x = 1-2\sin^2x$ is the way to go.
$\endgroup$ $\begingroup$$$\frac{2\sin x\cos x}{1-(1-2\sin^2x)}=\frac{\cos x}{\sin x}.$$
$\endgroup$
