# Are all values of $x$ solutions for $e^{2\ln(\sin(x))} = 1 - e^{2\ln(\cos(x))}$ in $\mathbb R$?

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.

• What exactly do you mean by $\ln*\sin(x)$? Does it refer to $\ln(\sin(x))$ or what is the meaning of this notation? – mrtaurho Mar 13 at 17:52
• sorry, my bad. Yes. Will fix equation – Peter Parada Mar 13 at 17:53
• 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. – 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 $$\boxed{x\in \color{red}{\bigcup_{k\in \mathbb{Z}} (0+2\pi k,{\pi\over 2}+2\pi k)}}$$

• Great. I was thinking the same. Does that mean solution on Wolfram is wrong? – Peter Parada Mar 13 at 18:04
• (+1) For myself I totally forgot to think about the existence of the natural logarithm as it is only assured for $\sin(x),\cos(x)>0$. – mrtaurho Mar 13 at 18:04
• @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. – Aqua Mar 13 at 18:07
• Thank you. I will write them for further clarification. – Peter Parada Mar 13 at 18:07
• @PeterParada: I think Wolfram implicitly assumes that $x$ is in the domain; you might need to check documentation to see if this is standard. – 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.