Q.Find the principal value of $\sin^{-1}\dfrac{1}{2}$

No trigonometric function is one-one on its entire natural domain, since periodicity means infinitely many angles share the same function value — so none can be inverted over all of $\mathbb{R}$ (or $\mathbb{R}$ minus its excluded points). Examining each function's graph, however, shows a specific restricted interval on which it becomes one-one and onto, and it is only on that restricted interval that an inverse is defined: sine on $\left[-\frac\pi2,\frac\pi2\right]$ gives $\sin^{-1}:[-1,1]\to\left[-\frac\pi2,\frac\pi2\right]$; cosine on $[0,\pi]$ gives $\cos^{-1}:[-1,1]\to[0,\pi]$; tangent on $\left(-\frac\pi2,\frac\pi2\right)$ gives $\tan^{-1}:\mathbb{R}\to\left(-\frac\pi2,\frac\pi2\right)$; and similarly restricted domains give $\text{cosec}^{-1}$, $\sec^{-1}$, and $\cot^{-1}$. For $x$ in the domain of an inverse function, the unique angle $\theta$ in its restricted range with (for example) $\sin\theta=x$ is called the principal value of $\sin^{-1}x$ — even though other angles outside the range also have sine $x$, only the one inside the agreed restricted interval counts as the principal value, which is why $\sin^{-1}(\sin y)=y$ only holds when $y$ already lies in $\left[-\frac\pi2,\frac\pi2\right]$, and must otherwise first be reduced to an equivalent angle that does.