Finding the Principal Value of sin⁻¹(sin(2π/3))

The question asks us to find the principal value of the expression \({\sin ^{ - 1}}\left( {\sin \frac{{2{\rm{\pi }}}}{3}} \right)\). To solve this, we need to understand the concept of the principal value branch of the inverse sine function.

Understanding the Inverse Sine Function's Principal Range

The inverse sine function, denoted as \({\sin ^{ - 1}}(x)\) or \(\arcsin(x)\), is defined for values of \(x\) in the domain \([-1, 1]\). The principal value branch of \({\sin ^{ - 1}}(x)\) has a range of \(\left[ { - \frac{\pi }{2}, \frac{\pi }{2}} \right]\). This means that for \(y = {\sin ^{ - 1}}(x)\), the value of \(y\) must lie within this interval.

The property \({\sin ^{ - 1}}\left( {\sin x} \right) = x\) is only true if \(x\) is within the principal value range of \({\sin ^{ - 1}}\), i.e., \(x \in \left[ { - \frac{\pi }{2}, \frac{\pi }{2}} \right]\).

Evaluating the Given Expression

In our problem, the expression is \({\sin ^{ - 1}}\left( {\sin \frac{{2{\rm{\pi }}}}{3}} \right)\). Here, the value of \(x\) is \(\frac{{2{\rm{\pi }}}}{3}\).

Let's check if \(\frac{{2{\rm{\pi }}}}{3}\) is within the principal value range \(\left[ { - \frac{\pi }{2}, \frac{\pi }{2}} \right]\).

• \(\frac{{2{\rm{\pi }}}}{3}\) in degrees is \(\frac{2 \times 180^\circ}{3} = 2 \times 60^\circ = 120^\circ\).
• The range \(\left[ { - \frac{\pi }{2}, \frac{\pi }{2}} \right]\) in degrees is \(\left[ -90^\circ, 90^\circ \right]\).

Since \(120^\circ\) is not within the interval \(\left[ -90^\circ, 90^\circ \right]\), we cannot directly use the property \({\sin ^{ - 1}}\left( {\sin x} \right) = x\).

Using Trigonometric Identities to Find an Equivalent Angle

We need to find an angle \(\theta\) such that \(\sin \theta = \sin \frac{{2{\rm{\pi }}}}{3}\) and \(\theta\) is within the principal value range \(\left[ { - \frac{\pi }{2}, \frac{\pi }{2}} \right]\).

We know the trigonometric identity: \(\sin (\pi - x) = \sin x\).

Using this identity, we can write:

\(\sin \frac{{2{\rm{\pi }}}}{3} = \sin \left( \pi - \frac{{2{\rm{\pi }}}}{3} \right)\)

Now, let's calculate the value inside the parenthesis:

\(\pi - \frac{{2{\rm{\pi }}}}{3} = \frac{3\pi - 2\pi}{3} = \frac{\pi}{3}\)

So, \(\sin \frac{{2{\rm{\pi }}}}{3} = \sin \frac{{\pi}}{3}\).

Verifying the Angle is in the Principal Range

Now, we consider the expression \({\sin ^{ - 1}}\left( {\sin \frac{{\pi}}{3}} \right)\).

Let's check if \(\frac{{\pi}}{3}\) is within the principal value range \(\left[ { - \frac{\pi }{2}, \frac{\pi }{2}} \right]\).

• \(\frac{{\pi}}{3}\) in degrees is \(\frac{180^\circ}{3} = 60^\circ\).
• The range \(\left[ { - \frac{\pi }{2}, \frac{\pi }{2}} \right]\) in degrees is \(\left[ -90^\circ, 90^\circ \right]\).

Since \(60^\circ\) is within the interval \(\left[ -90^\circ, 90^\circ \right]\), we can now use the property \({\sin ^{ - 1}}\left( {\sin x} \right) = x\).

Final Calculation

Therefore, \({\sin ^{ - 1}}\left( {\sin \frac{{2{\rm{\pi }}}}{3}} \right) = {\sin ^{ - 1}}\left( {\sin \frac{{\pi}}{3}} \right) = \frac{{\pi}}{3}\).

The principal value of \({\sin ^{ - 1}}\left( {\sin \frac{{2{\rm{\pi }}}}{3}} \right)\) is \(\frac{{\pi}}{3}\).

Comparing with the options, the correct answer is \(\frac{{\pi}}{3}\).

Revision Table: Inverse Trigonometric Functions

Additional Information: Properties of Inverse Sine

Here are some important properties related to the inverse sine function:

• Domain: \([-1, 1]\)
• Range (Principal Value): \(\left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]\)
• Identity: \({\sin ^{ - 1}}(\sin x) = x\), only if \(x \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]\). If \(x\) is outside this range, we need to find an equivalent angle within the range using trigonometric identities like \(\sin x = \sin (\pi - x)\), \(\sin x = \sin (2\pi + x)\), etc., or by understanding the graph of \(\sin x\).
• Identity: \(\sin ({\sin ^{ - 1}}x) = x\), for all \(x \in [-1, 1]\).
• Odd function: \({\sin ^{ - 1}}(-x) = -{\sin ^{ - 1}}(x)\) for \(x \in [-1, 1]\).

Understanding the principal value ranges for all inverse trigonometric functions is crucial for solving problems involving them.