Prove this identity: $\frac{2\sin^4x+\cos^2x-2\cos^4x}{3\sin^2x-1} =1$

Question

I am stuck on this identity $$\frac{2\sin^4x+\cos^2x-2\cos^4x}{3\sin^2x-1} =1$$

I began working on the left side trying to get things to cancel out or equal one by the Pythagorean identities. I am stuck and can't get it to reduce anymore.

Answer 1

Hint: Use the fact that $$\sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x) = \cdots$$ You'll see that things cancel nicely.

Answer 2

A variation is to apply the identity $ \ \sin^4 x \ - \ \cos^4 \ = \ \sin^2 x \ - \ \cos^2 x \ $ , as Clive Newstead shows (through the Pythagorean Identity), to write

$$ \frac{2 \ (\sin^2 x \ - \cos^2 x) \ + \ \cos^2 x}{\sin^2 x \ + \ (2 \sin^2x - 1) } \ = \ \frac{2 \ (- \cos \ 2x) \ + \ \frac{1}{2} (1 + \cos \ 2x)}{\frac{1}{2} (1 - \cos \ 2x) \ + \ (- \cos \ 2x) } \ \ , $$

by applying the "double-angle" formulas for cosine and the "sine-squared" and "cosine-squared" identities.

Answer 3

Let's take this one step at a time. First, move the denominator to the right side. 1) $$ 2sin^4(x) + cos^2(x) - 2cos^4(x) = 3sin^2(x) - 1 $$ Use the identity $ sin^2(x) = 1-cos^2(x) $ to rewrite $ sin^4(x) = (1-cos^2(x))^2$.
2) $$ 2(1-cos^2(x))^2 + cos^2(x) - 2cos^4(x) = 3sin^2(x) - 1 $$ Now foil out $2(1-cos^2(x))^2 = 2(1 - 2cos^2(x) + cos^4(x))$ and distribute the 2. 3) $$ 2-4cos^2(x) + 2cos^4(x) + cos^2(x) - 2cos^4(x) = 3sin^2(x) - 1$$ Now there are three things we can do here First we have $ 2cos^4(x) $ and $ -2cos^4(x) $ so those will cancel out. Second we have $ -4cos^2(x) + cos^2(x) = -3cos^2(x)$. And last we can move the 2 from the left side of the equation to the right. 4) $$ -3cos^2(x) = 3sin^2(x) - 3 $$ Now factor out the 3 and it cancels out 5) $$-cos^2(x) = sin^2(x) - 1 $$ And just rearrange it and we get 6) $$ 1 = sin^2(x) + cos^2(x) $$

Hope that helps!

Answer 4

You want to prove $$\frac{2\sin^4x+\cos^2x-2\cos^4x}{3\sin^2x-1} =1$$ which is equivalent to $$2\sin^4x+\cos^2x-2\cos^4x=3\sin^2x-1$$ which is in turn equivalent to $$(2\sin^4x+\cos^2x-2\cos^4x)-(3\sin^2x-1)=0.$$ Consider the function $f$ with $f(x)=(2\sin^4x+\cos^2x-2\cos^4x)-(3\sin^2x-1)$. We can compute the derivative: $$ f'(x)=8\sin^3 x\,\cos x-2\cos x\sin x+8\cos^3x\sin x-6\sin x\cos x $$ and we can pull out the common factor $8\sin x\cos x$ to rewrite this as $$f(x)=-8\sin x\cos x(\sin^2x+\cos^2x-1)$$ which is clearly zero, so that $f$ is constant. As $f(0)=0$, as one can check at once, then $f(x)=0$ for all $x$, which proves what we wanted.