Let $\theta=\sin^{-1}\frac{1}{2}$ We know,$\sin(\pi-\theta)=\sin \theta$ or, $$\sin(\pi-\sin^{-1}\frac{1}{2})=\sin(\sin^{-1}\frac{1}{2})$$ or, $$\pi-\sin^{-1}\frac{1}{2}=\sin^{-1}\frac{1}{2}$$ or $$2\sin^{-1}\frac{1}{2}=\pi$$ or $$\sin^{-1}\frac{1}{2}=\frac{\pi}{2}$$ or $$\frac{1}{2}=\sin \frac{\pi}{2}$$ or $$\frac{1}{2}=1$$ I don't understand what's wrong with the argument.I know I messed up the principal values but can't yet find the error.
$\begingroup$
$\endgroup$
9
-
1$\begingroup$ The error is just after your second or. if $\sin a = \sin b $ then $ a=b+2k\pi or a = \pi-b +2k\pi $ $\endgroup$– hamam_AbdallahJul 22, 2021 at 20:17
-
$\begingroup$ But if the domain of $\sin$ is from $-\frac{\pi}{2} to \frac{\pi}{2}$,then isn't it okay to cancel the values? $\endgroup$– a_i_rJul 22, 2021 at 20:20
-
1$\begingroup$ $\sin x=\sin y$ does not mean $x=y.$ You've just used a theorem that proves that, since $\sin(x)=\sin(\pi -x).$ $\endgroup$– Thomas AndrewsJul 22, 2021 at 20:22
-
$\begingroup$ @a_i_r but $ \pi -\sin^{-1}1/2$ is in $[\pi/2 ,3\pi/2]$ $\endgroup$– hamam_AbdallahJul 22, 2021 at 20:23
-
1$\begingroup$ @a_i_r To be clear, if you're confused about the answers there is an error in the statement $\sin(x)=\sin(y)\therefore x=y$. In your case, although we know $\sin(\pi-\theta)=\sin(\theta)$, do note that $\pi-\theta\neq\theta$ unless $\theta=\frac{\pi}{2}$ $\endgroup$– FShrikeJul 22, 2021 at 21:01
|
Show 4 more comments
3 Answers
$\begingroup$
$\endgroup$
It is true that if $x,y\in\left[-\frac\pi2,\frac\pi2\right]$ and $\sin(x)=\sin(y),$ then $x=y.$
It is also true that $\sin(\pi -x)=\sin(x)$ for any $x.$
These two together show, with your argument, that if $x$ and $\pi-x$ are both in the range $\left[-\frac\pi2,\frac\pi2\right]$ then $x=\frac\pi2.$
But if $x<\frac{\pi}{2}.$ then $\pi-x>\frac{\pi}{2}.$ So this is not as big a result as one might initially think.
answered Jul 22, 2021 at 20:54
$\begingroup$
$\endgroup$
The principal solution to $\theta=\sin^{-1}(\frac 12)$ is $\theta=\frac\pi 6$
Note that $\pi-\frac\pi 6=\frac{5\pi}{6}\neq \frac \pi 6$ (while $\sin(\frac{5\pi}{6})=\frac12$ still holds)
Indeed, the quality $f(x)=f(y)\implies x=y$ defines an injective, or one-to-one, function, which $\sin$ is not.
answered Jul 22, 2021 at 20:25
$\begingroup$
$\endgroup$
Your mistake is that $\require{cancel} \sin x=\sin y \implies x \cancel= y$ . This might be of some use. Take for example $ \sin 135 = \sin 405$ but here $135\cancel=405$.