3
$\begingroup$

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!!

$\endgroup$
  • 2
    $\begingroup$ For real $x$ the equation $\sin^4x-\cos^4x=1$ can hold only when $\sin^2x=1$ and $\cos^2x=0$. And you divided by $\cos^2x$... $\endgroup$ – Jyrki Lahtonen May 27 '16 at 6:42
  • $\begingroup$ @JyrkiLahtonen You should have made it an answer instead of a comment. $\endgroup$ – Marcelo Ventura May 27 '16 at 8:16
5
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $n \in \mathbb{Z}$.

$\endgroup$
2
$\begingroup$

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

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

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

$\endgroup$
  • $\begingroup$ you likely want to point out that this forces $\cos x = 0$ and so you cannot divide by it... $\endgroup$ – gt6989b May 27 '16 at 6:49
  • 1
    $\begingroup$ @gt6989b, Where have I divided by zero? $\endgroup$ – lab bhattacharjee May 27 '16 at 6:52
  • $\begingroup$ This is not an answer to the question "what is the fallacy...?" $\endgroup$ – Jyrki Lahtonen May 27 '16 at 9:19
  • $\begingroup$ you did not, just the OP does, and that is the problem with his approach... $\endgroup$ – gt6989b May 27 '16 at 18:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.