I need to prove that this is an identity:
$$\frac{\sin 2x - \cos x}{4\sin^2x -1} = \frac{\sin^2x+\cos x+\cos^2x}{2\sin x +1} $$
Here's what I'm confused on:
I noticed that the expression in the denominator for the first expression ($4sin^2(x) -1)$ looks like a double angle identity, but it also looks like a difference of squares. I know I can solve this as a difference of squares. Am I correct that this is a double angle identity, and if so, how can I solve it?
I could rearrange the numerator for the second expression ($\sin^2x+\cos x+\cos^2x$) to sub in $1$ for $\sin^2\theta+\cos^2\theta$ (Pythagorean identity, where $\sin^2\theta+\cos^2\theta=1$) By doing so, I'd get $\cos x + 1$. But, this must be incorrect, for the final answer is $\dfrac{\cos x}{2\sin x-1} = \dfrac{\cos x}{2\sin x-1}$. That means the cosine must've been multiplied by $1$. How is that so?