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

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)

• Are you sure you recorded the problem correctly? Feb 22, 2019 at 3:34
• @Bladewood I copied this problem off from an illegible picture of math notes my friend has sent me. If it helps I can attach that photo to this problem. Feb 22, 2019 at 3:46
• Okay, that changes things. This looks like a completely different problem to me: the numerator of the RHS is $\sin^2 x \cos x + \cos^3 x$, which the second line follows by factoring out $\cos x$ (although I'm not sure how the rest of it follows). Feb 22, 2019 at 4:11

$$\frac{\sin(2x) - \cos x}{4\sin^2(x) -1 }= \frac{2\sin x\cos x - \cos x}{(2\sin x -1)(2\sin x +1)} = \frac{\cos x}{2\sin x +1} \color{red}{\neq} \frac{1+\cos x}{2\sin x +1}$$
When $$x=0$$ the left side equals 1 and the right side equals 2, so it is not an identity.
$$\sin(2x)=2\sin (x)\cos (x)$$. 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.
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 $$(\cos x)(\cos^2 x+\sin^2 x)$$.
• ¿Que? There are definitely no solutions as they require $\cos(x)=\cos(x)+1$ Feb 22, 2019 at 4:14