I have a question that says to find the exact value of $\sin^{-1} (1/\sqrt{2})$
For the answer it shows $\frac{\pi}{4} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Where does that last part in [] comes from? What is it? $\pi/4$ is $45$ degrees, I understand how they get that part but not the other.
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$\begingroup$What you describe seems to be $$ \frac{\pi}{4} \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right] $$
$x \in A$ means that $x$ is a member of the set $A$.
$[a,b]$, "the closed interval $a$, $b$" means the set of real numbers between, or equal to one of, $a$ and $b$, i.e. $$ [a,b] = \{ x \in \mathbb{R} : a \leqslant x \leqslant b \}. $$
$\endgroup$ 1 $\begingroup$By definition of sine inverse, $\sin^{-1}(\frac{1}{\sqrt{2}}) = \theta \iff \sin \theta = \dfrac{1}{\sqrt{2}}$ and $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Only one angle that satisfies this: $\theta = \dfrac{\pi}{4}$ or $45^{\circ}$.
The part $\theta \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ means that $\theta$ is greater than $-\frac{\pi}{2}$, and is less than $\frac{\pi}{2}$.
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