What is the fallacy of this trigonometrical proof that $1=-1?$

Question

I have this equation which I solved-

$$\sin^4x-\cos^4x=1$$ $$\implies -\cos^4x=1-\sin^4x$$ $$\implies-\cos^4x=(1+\sin^2x)(1-\sin^2x)$$ $$\implies-\cos^4x=(1+\sin^2x)\cos^2x$$ $$\implies-\cos^2x=(1+\sin^2x)$$ $$\implies-(\cos^2x+\sin^2x)=1$$ $$\implies-1=1$$

I came across this thing while solving a problem and I am at total loss why this is coming.Where am I wrong?

Thanks for any help!!

Answer 1

Your inference \begin{align} & & -\cos^4x &= (1+\sin^2x)\cos^2x \\ &\implies & -\cos^2x &= (1+\sin^2x) \end{align} is false. The valid inference is \begin{align} & & -\cos^4x &= (1+\sin^2x)\cos^2x \\ &\implies & -\cos^2x &= (1+\sin^2x) \quad \text{ or }\quad \cos^2 x = 0 \end{align} This accounts for the possibility that you have divided by zero. Subsequently, the left choice yields no solutions and the right choice yields the solution set of the equality in the first line of your problem.

Answer 2

The fault lies in these steps:

$$\implies-\cos^4x=(1+\sin^2x)\cos^2x$$ $$\implies-\cos^2x=(1+\sin^2x)$$

The cancellation of $\cos^2 x$ on both sides implies the assumption that $\cos x \not =0$ which is not logical since in the first step above, the L.H.s is negative, whereas the R.H.S. is positive.

So the only solution of the equation is $\cos x=0$ or in other words, $x=2n\pi \pm \frac{\pi}{2}$ where $n \in \mathbb{Z}$.

Answer 3

$$\sin^4x-\cos^4x=-(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)=-\cos2x$$

$$\iff\cos2x=-1=\cos\pi$$

$$\implies2x=(2m+1)\pi$$ where $m$ is any integer