Find y if y = arcsin(sqrt(3)/2).

Step-by-step solution

1. 1

   State what the inverse sine returns. $\sin^{-1}(a)$ is the unique angle $y$ with

   $\sin y=a\qquad\text{and}\qquad -\frac{\pi}{2}\le y\le\frac{\pi}{2}$

   Restricting the range is what makes the inverse a function at all; without it there would be infinitely many answers.

2. 2

   Check the input is in range. The domain of $\sin^{-1}$ is $[-1,1]$, and $\dfrac{\sqrt3}{2}\approx 0.8660$ lies inside it, so an answer exists.

3. 3

   Recall the special angle. From the $30$-$60$-$90$ triangle (or the unit circle),

   $\sin\frac{\pi}{3}=\frac{\sqrt3}{2}$

   The three values worth memorising are $\sin\frac\pi6=\frac12$, $\sin\frac\pi4=\frac{\sqrt2}{2}$, $\sin\frac\pi3=\frac{\sqrt3}{2}$ — increasing numerators under the same denominator.

4. 4

   Confirm the candidate lies in the principal range. Since $0<\dfrac{\pi}{3}<\dfrac{\pi}{2}$, the angle $\tfrac\pi3$ is admissible, so

   $y=\frac{\pi}{3}=60^{\circ}$

5. 5

   Note the answer that is deliberately excluded. $\sin\dfrac{2\pi}{3}$ is also $\dfrac{\sqrt3}{2}$, and so is $\sin\left(\frac{\pi}{3}+2\pi k\right)$ for every integer $k$. All of these solve the equation $\sin y=\frac{\sqrt3}{2}$, but only $\tfrac\pi3$ is the value of $\sin^{-1}$. Confusing "solve $\sin y=a$" with "evaluate $\sin^{-1}a$" is the standard error here.