Question

Evaluate the principal value of the inverse trigonometric expression: \( \sin^{-1}\left(\sin\frac{2\pi}{3}\right) \)

• A. \( \frac{\pi}{3} \)
• B. \( \frac{2\pi}{3} \)
• C. \( -\frac{\pi}{3} \)
• D. \( \frac{\pi}{6} \)

Hint:

On a unit circle, the sine of an angle is just the y-coordinate of the point at that angle. Since sin inverse only outputs angles between minus pi over 2 and pi over 2, first find where 2 pi over 3 lands on the circle, then look for its mirror point in that range sharing the same y-coordinate; that mirror angle is the answer.

Answer 1

The Correct Option is A

Concept: The identity \( \sin^{-1}(\sin\theta) = \theta \) holds true if and only if the angle \( \theta \) falls inside the restricted principal value branch of the inverse sine function: \[ \theta \in \left[-\frac{\pi}{2}, \, \frac{\pi}{2}\right] \] If the given angle falls outside this boundary domain, you must rewrite it using standard trigonometric quadrant identity properties first.

Step 1: Verify if the given angle falls within the principal branch region.
The given angle is \( \theta = \frac{2\pi}{3} \) (which is \( 120^\circ \)). This values sits outside the eligible principal boundary range of \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) (\([-90^\circ, 90^\circ]\)), meaning we cannot cancel the operators directly. \[ \frac{2\pi}{3} \notin \left[-\frac{\pi}{2}, \, \frac{\pi}{2}\right] \]

Step 2: Use trigonometric identities to map the angle into the principal quadrant.
Since the sine function is positive in the second quadrant, we use the supplementary identity rule \( \sin(\pi - x) = \sin x \): \[ \sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) \]

Step 3: Evaluate the principal inverse value expression.
Substitute our rewritten equivalent expression back into the function: \[ \sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \sin^{-1}\left(\sin\frac{\pi}{3}\right) \] Since \( \frac{\pi}{3} \) (\( 60^\circ \)) sits safely inside the valid principal interval, the operators cancel out directly: \[ \sin^{-1}\left(\sin\frac{\pi}{3}\right) = \frac{\pi}{3} \]

Answer 2

Concept:

• On the unit circle, the sine of an angle equals the $y$-coordinate of the point at that angle.
• Since the principal branch of $\sin^{-1}$ only accepts angles from $-\pi/2$ to $\pi/2$ (the right half of the unit circle), any given angle outside this range can be replaced by its mirror-image point in the right half that has the same $y$-coordinate.

Step 1: Locate the point on the unit circle at angle $2\pi/3$.
$2\pi/3$ radians is $120^\circ$, which lies in the second quadrant. The coordinates of this point are:
$\left( \cos\dfrac{2\pi}{3}, \sin\dfrac{2\pi}{3} \right) = \left( -\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} \right)$

Step 2: Note that the principal branch of $\sin^{-1}$ only covers the right half of the circle.
The principal value range of $\sin^{-1}$ is $[-\pi/2, \pi/2]$, which corresponds to points on the right half of the unit circle, where the $x$-coordinate is non-negative. Since $2\pi/3$ lies in the left half, its point has $x = -1/2 < 0$, so it is not itself the answer.

Step 3: Reflect the point to the right half of the circle.
Reflecting $\left( -\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} \right)$ across the $y$-axis gives the point $\left( \dfrac{1}{2}, \dfrac{\sqrt{3}}{2} \right)$, which has the same $y$-coordinate, so the same sine value. This point lies in the right half of the circle, at angle:
$\dfrac{\pi}{3}$

Step 4: Confirm this angle gives the required sine value.
$\sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2} = \sin\dfrac{2\pi}{3}$, and $\dfrac{\pi}{3}$ lies inside $[-\pi/2, \pi/2]$, so it is the unique principal value.

Final Answer: $\sin^{-1}\left( \sin \dfrac{2\pi}{3} \right) = \dfrac{\pi}{3}$