Proving $\frac{\sin 2x - \cos x}{4\sin^2x -1} = \frac{\sin^2x+\cos x+\cos^2x}{2\sin x +1} $

Question

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?

(Here's the picture of the problem)

Answer 1

You cannot prove it because it is false:

$$\frac{\sin(2x) - \cos x}{4\sin^2(x) -1 }= \frac{2\sin x\cos x - \cos x}{(2\sin x -1)(2\sin x +1)} = \frac{\cos x}{2\sin x +1} \color{red}{\neq} \frac{1+\cos x}{2\sin x +1}$$

Answer 2

When $x=0$ the left side equals 1 and the right side equals 2, so it is not an identity.

Answer 3

$\sin(2x)=2\sin (x)\cos (x)$. Therefore you are solving: $$\frac{\cos (x)}{2\sin(x)+1}=\frac{\cos(x)+1}{2\sin(x)+1}$$ Which leads to the nonsense equation: $$\cos(x)=\cos(x)+1$$ which certainly has no solutions.

Answer 4

As other have pointed out, the equation is not an identity. To directly address your confusions, $4\sin^2(x)-1$ is similar to the double-angle identity $\cos(2x)=1-2\sin^2(x)$, from which we can derive the identity $4\sin^2(x)-1=-2\cos(2x)+1$. However, it doesn't seem using this substitution helps, and only complicates it by changing the easier to deal with $\sin^2(x)$ term into the more difficult $\cos(2x)$. Since the equation is not a identity, $\dfrac{\cos x}{2\sin x-1} = \dfrac{\cos x}{2\sin x-1}$ does not follow from the equation. The Pythagorean identity is actually the correct substitution to make. Your friend has made a mistake in the RHS in the second line, incorrectly factoring $\sin^2(x)+\cos(x)+\cos^2(x)$ into $(\cos x)(\cos^2 x+\sin^2 x)$.