Using the identity $\sin(\theta)=\sin(180°-\theta)$, I obtained some contradictory results. If we let $\theta=70°$, then we get that \begin{align} \sin(70°) &= \sin(180° - 70°) = 0.9396926208 \\[5pt] \sin^{-1}(\sin(70°)) &= \sin^{-1}(\sin(180° - 70°)) \\[5pt] 70° &= 180° - 70° \\[5pt] 140° &= 180° \, .\\ \end{align} The identity also seems to imply that $\theta=90°$: \begin{align} \sin^{-1}(\sin(\theta))&=\sin^{-1}(\sin(180°-\theta)) \\[5pt] \theta&=180°-\theta \\[5pt] \theta&=90° \end{align} Both of these results seem mistaken, but I don't know where I went wrong.

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    $\begingroup$ You cannot conclude that $70º = 180º - 70º$ as $\arcsin(\sin x)$ is only one-to-one in the domain $[-\pi/2, \pi/2]$. $\endgroup$
    – Toby Mak
    Jul 22 at 8:34
  • $\begingroup$ Do you know the relation between trigonometric ratios of $\theta$ and $180^\circ-\theta$? $\endgroup$ Jul 22 at 8:36
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    $\begingroup$ +1 to your question for good work shown, taking your question's score from (-3) to (-2). The fact that your work had an analytical mistake (as indicated by the other comments) does not imply that your posting is low quality. My only explanation for why (others) have downvoted or voted to close your question is that there is no quality control on the behavior of mathSE reviewers. $\endgroup$ Jul 22 at 10:36
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    $\begingroup$ @TobyMak: Your comment does really get to the heart of the issue. However, I would suggest that you use degrees rather than radians, as Mohammad might not be familiar with radians. $\endgroup$
    – Joe
    Jul 22 at 11:29
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    $\begingroup$ Does this answer your question? Why aren't the graphs of $\sin(\arcsin x)$ and $\arcsin(\sin x)$ the same? $\endgroup$
    – Toby Mak
    Jul 22 at 11:31

First of all, $\sin(x)=\sin(180-x)$ is an identity, meaning that it is true for all values of $x$. So it does not make sense to "solve for $x$"—that would be akin to trying to solve for $x$ in the identity $(x+1)^2=x^2+2x+1$.

By definition, the inverse function $f^{-1}$ satisfies $f^{-1}(f(x))=x$ for all $x$ in the domain of $f$. The issue is that $\sin^{-1}$ is technically not the inverse of $\sin$, but rather the inverse of $\sin$ on a restricted domain, i.e. it is the inverse of the function $f(x)=\sin(x)$, where $\color{red}{-90\le x\le 90}$. So the identity $$ \sin^{-1}(\sin(x))=x $$ only applies when $-90\le x \le 90$. If $x$ is not in this interval, then $\sin^{-1}(\sin(x))\neq x$. In particular, if $x=180-70$, then $\sin^{-1}(\sin(180-70))\neq 180-70$. For much the same reason, from the identity $\sin^{-1}(\sin(x))=\sin^{-1}(\sin(180-x))$ you cannot conclude that $x=180-x$, as you do not know that $x$ and $180-x$ both lie in the interval $[-90,90]$.

  • $\begingroup$ From what you have shown, I believe I won't be able to use the sin$^{-1}$ to remove the sin from sin$(180-x)$, right? Is there a way for me to still get the value of $x$ from sin $x$ even when $x > 90º$? Also, I obtained $140º = 180º$ by adding $70º$ to both sides of, $70º = 180º - 70º$, to remove $-70º$. $\endgroup$ Jul 22 at 12:36
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    $\begingroup$ @Mohammadmuazzamali: Yes, $\sin^{-1}(\sin(180-x))\neq 180-x$ unless $-90\le180-x\le90$. As for whether there is a way of getting the value of $x$ from $\sin x$, the answer is no, sadly. If $-1\le c \le 1$, then the equation $\sin x = c$ has infinitely many solutions. So there is no way of "undoing" this equation to find $x$, unless we are told that $x$ falls into a specific interval like $[-90,90]$ or $[90,180]$. By the way, once you got that $70=180-70$, this was already an error, and so it doesn't really matter how you manipulate this equation—you're very likely to end up with more nonsense. $\endgroup$
    – Joe
    Jul 22 at 14:52

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