# Simplify $\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$

Simplify: $$\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$$

After the substitution as $$\cos(x)=a$$ and $$\sin(x)=b$$, $$(a^2+b^2=1)$$, the expression becomes $$\frac{4(a^2-b^2)^2-4a^2+3b^2}{4b^2-4a^2b^2}=\frac{4a^4-8a^2b^2+4b^4-4a^2+3b^2}{4b^4}=\bigg(\frac{a^2}{b^2}-1\bigg)^2-\frac{4a^2-3b^2}{4b^4}$$But I don't think I got anything useful... Any help is appreciated.

\begin{align}\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}&=\frac{4(1-2\sin^2x)^2-4(1-\sin^2x)+3\sin^2x}{4\sin^2x-(2\sin x\cos x)^2}\\&=\frac{4-16\sin^2x+16\sin^4x-4+4\sin^2x+3\sin^2x}{4\sin^4x}\\&=\frac{16\sin^4x-9\sin^2x}{4\sin^4x}\\&=4-\frac94\csc^2x\end{align}

Let us express everything in term of $$s:=\sin x$$:

$$\frac{4(1-2s^2)^2-4(1-s^2)+3s^2}{4s^2-4s^2(1-s^2)}=\frac{16s^4-9s^2}{4s^4}=4-\frac9{4s^2}.$$

• I think you omitted the square $^2$ on $\cos^2(2x)$ Commented Jun 27, 2020 at 10:00
• @VIVID: yes, I just noticed. Now fixed. Cheers.
– user65203
Commented Jun 27, 2020 at 10:00

Another way is to express everything in terms of $$\cos 2x$$: $$\frac{4\cos^22x -3(\cos^2x-\sin^2x)-\cos^2x}{4\cdot\frac{1-\cos2x}{2} -(1-\cos^22x)} \\=\frac{4\cos^22x-3\cos 2x -\frac{1+\cos 2x}{2}}{2(1-\cos 2x)+\cos^22x-1}\\=\frac{8\cos^22x-7\cos 2x-1}{2(1-\cos 2x)^2}\\=\frac{(8\cos 2x+1)(\cos 2x-1)}{2(\cos 2x -1)^2}\\=\frac{8\cos 2x+1}{2\cos 2x-2}$$

So we have : $$\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$$ First of all, by using double-angles formulas we can get this: $$\cos ^2\left(x\right) = \frac{1+\cos\left(2x\right)}{2}$$ $$\sin ^2\left(x\right) = \frac{1-\cos\left(2x\right)}{2}$$ So $$4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right) = \frac{8\cos ^2\left(2x\right)-7\cos \left(2x\right)-1}{2}$$ which equals to $$\frac{(cos\left(2x\right) -1)(8\cos \left(2x\right)+1)}{2}$$ Lets look at the denominator $${4\sin ^2\left(x\right)-\sin ^2\left(2x\right) = \cos^2 \left(2x\right) - 2\cos \left(2x\right) + 1}$$ Since: $$4\sin ^2\left(x\right) = 2 -2\cos \left(2x\right)$$ $$\sin^2\left(2x\right) = 1 - \cos^2 \left(2x\right)$$ So we get: $$(\cos^2 \left(2x\right) - 2\cos \left(2x\right) + 1) = (\cos \left(2x\right) - 1)^2$$ Combining these equations, our first equation becomes : $$\frac{(cos\left(2x\right) -1)(8\cos \left(2x\right)+1)}{2(\cos \left(2x\right) - 1)^2}$$ Which can be reduced to: $$\frac{8\cos \left(2x\right)+1}{2\cos \left(2x\right) - 2}$$ By using half angle formulas, we have : $$8\cos\left(2x\right) = 8 - 16\sin^2\left(x\right)$$ $$2\cos\left(2x\right) = 2 - 4\sin^2\left(x\right)$$ So our final result is : $$\frac{8\cos \left(2x\right)+1}{2\cos \left(2x\right) - 2 } = \frac{16\sin^2\left(x\right) - 9}{4\sin^2\left(x\right)}$$.