Does all values of $x$ in $\mathbb R$ satisfy equation: $$e^{2\ln(\sin(x))} = 1 - e^{2\ln(\cos(x))}$$

I am asking this, because by checking WolframAlpha solution there is an answer: (all values for $x$ are solutions over reals), but we know that $\ln(0)$ is undefined, same for negative numbers.

Wolfram Alpha solution

Therefore I assume in $\mathbb R$, zero and negative numbers doesn't satisfy this equation.

  • $\begingroup$ What exactly do you mean by $\ln*\sin(x)$? Does it refer to $\ln(\sin(x))$ or what is the meaning of this notation? $\endgroup$ – mrtaurho Mar 13 at 17:52
  • $\begingroup$ sorry, my bad. Yes. Will fix equation $\endgroup$ – Peter Parada Mar 13 at 17:53
  • 1
    $\begingroup$ Wolfram Alpha will always give explicitly Wrong Answer for some input. Any real mathematician will always be able to easily find a limit that exists but WA says it doesn't, or vice versa. Whenever someone reports a wrong answer, the programmers just hack a 'fix' based on random guesswork rather than mathematics. Bottom line: Don't trust WA to be correct even on very simple questions. $\endgroup$ – user21820 Mar 27 at 6:34

$$1=e^{2\ln\sin(x)} +e^{2\ln \cos(x)}= e^{\ln\sin^2(x)}+e^{\ln \cos^2(x)} = $$

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

So this is true for all $x$ such that $\sin x>0$ and $\cos x>0$ (because then is $\ln $ defined).

So $$\boxed{x\in \color{red}{\bigcup_{k\in \mathbb{Z}} (0+2\pi k,{\pi\over 2}+2\pi k)}}$$

  • $\begingroup$ Great. I was thinking the same. Does that mean solution on Wolfram is wrong? $\endgroup$ – Peter Parada Mar 13 at 18:04
  • 1
    $\begingroup$ (+1) For myself I totally forgot to think about the existence of the natural logarithm as it is only assured for $\sin(x),\cos(x)>0$. $\endgroup$ – mrtaurho Mar 13 at 18:04
  • $\begingroup$ @PeterParada I'm not sure what is Wolfram doing. It draw you a whole line for $f(x) =e^{\ln x}$ which is absurd to me. $\endgroup$ – Aqua Mar 13 at 18:07
  • $\begingroup$ Thank you. I will write them for further clarification. $\endgroup$ – Peter Parada Mar 13 at 18:07
  • $\begingroup$ @PeterParada: I think Wolfram implicitly assumes that $x$ is in the domain; you might need to check documentation to see if this is standard. $\endgroup$ – Clayton Mar 13 at 18:08

Note that : $$e^{2\ln \sin x} + e^{2 \ln \cos x} = 1 \Rightarrow e^{\ln (\sin x)^2} + e^{\ln (\cos x)^2} = 1 \Leftrightarrow \sin^2x + \cos^2x = 1 \rightarrow \text{true} \; \forall x \in \mathbb R$$ Restrictions apply so as the initial expression holds, so that narrows down the solution set. Note the usage of $\Rightarrow$ instead of an $\Leftrightarrow$ at the start.


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.