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The value of $\sin^{-1}\left(\dfrac{1}{2}\right)$ is:
A$\dfrac{\pi}{6}$
B$\dfrac{\pi}{2}$
C$\dfrac{\pi}{4}$
D$\dfrac{\pi}{3}$
Answer & Solution
Correct answer: A. $\dfrac{\pi}{6}$
$\sin\theta = \dfrac{1}{2}$ has the principal solution $\theta = \dfrac{\pi}{6}$ (30°), which lies in the standard range $[-\pi/2, \pi/2]$.
Memorise the three classic values: $\sin^{-1}\dfrac{1}{2} = \dfrac{\pi}{6}$, $\sin^{-1}\dfrac{1}{\sqrt 2} = \dfrac{\pi}{4}$, $\sin^{-1}\dfrac{\sqrt 3}{2} = \dfrac{\pi}{3}$.
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