Evaluate the Value of Expression Involving Inverse Trigonometry

We need to find the value of the given trigonometric expression: \({\cos ^{ - 1}}\left( {\cos \frac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\sin \frac{{5\pi }}{3}} \right)\). This problem involves evaluating inverse trigonometric functions.

Understanding the Principal Ranges of Inverse Trigonometric Functions

To evaluate expressions like \({\cos ^{ - 1}}(\cos \theta)\) and \({\sin ^{ - 1}}(\sin \theta)\), it's crucial to remember the principal value ranges of these functions:

• The principal value branch of \({\cos ^{ - 1}}(x)\) is \([0, \pi]\). This means \({\cos ^{ - 1}}(\cos \theta) = \theta\) only if \(0 \le \theta \le \pi\).
• The principal value branch of \({\sin ^{ - 1}}(x)\) is \([-\frac{\pi}{2}, \frac{\pi}{2}]\). This means \({\sin ^{ - 1}}(\sin \theta) = \theta\) only if \(-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}\).

In our expression, the angle is \(\frac{5\pi}{3}\). Let's check if this angle falls within the required ranges.

\(\frac{5\pi}{3}\) radians is equal to \(\frac{5 \times 180^\circ}{3} = 5 \times 60^\circ = 300^\circ\).

Clearly, \(300^\circ\) is not in \([0^\circ, 180^\circ]\) (for \({\cos ^{ - 1}}\)) and not in \([-90^\circ, 90^\circ]\) (for \({\sin ^{ - 1}}\)). Therefore, we cannot directly use the property \(f^{-1}(f(\theta)) = \theta\).

Simplifying the Term \({\cos ^{ - 1}}\left( {\cos \frac{{5\pi }}{3}} \right)\)

We need to rewrite \(\cos \frac{5\pi}{3}\) in terms of an angle \(\theta\) such that \(0 \le \theta \le \pi\), and \(\cos \theta = \cos \frac{5\pi}{3}\). The angle \(\frac{5\pi}{3}\) is in the 4th quadrant. We can write \(\frac{5\pi}{3}\) as \(2\pi - \frac{\pi}{3}\).

Using the property \(\cos(2\pi - \theta) = \cos \theta\), we have:

\(\cos \frac{5\pi}{3} = \cos \left(2\pi - \frac{\pi}{3}\right) = \cos \frac{\pi}{3}\)

Now, the expression becomes \({\cos ^{ - 1}}\left( {\cos \frac{\pi}{3}} \right)\). Since \(0 \le \frac{\pi}{3} \le \pi\), we can use the property \({\cos ^{ - 1}}(\cos \theta) = \theta\):

\({\cos ^{ - 1}}\left( {\cos \frac{\pi}{3}} \right) = \frac{\pi}{3}\)

So, the first part of the value of expression is \(\frac{\pi}{3}\).

Simplifying the Term \({\sin ^{ - 1}}\left( {\sin \frac{{5\pi }}{3}} \right)\)

Similarly, we need to rewrite \(\sin \frac{5\pi}{3}\) in terms of an angle \(\phi\) such that \(-\frac{\pi}{2} \le \phi \le \frac{\pi}{2}\), and \(\sin \phi = \sin \frac{5\pi}{3}\). The angle \(\frac{5\pi}{3}\) is in the 4th quadrant.

Using the property \(\sin(2\pi - \theta) = -\sin \theta\), we have:

\(\sin \frac{5\pi}{3} = \sin \left(2\pi - \frac{\pi}{3}\right) = -\sin \frac{\pi}{3}\)

Now, the expression becomes \({\sin ^{ - 1}}\left( {-\sin \frac{\pi}{3}} \right)\). We know that \(-\sin \theta = \sin (-\theta)\). So,

\({\sin ^{ - 1}}\left( {-\sin \frac{\pi}{3}} \right) = {\sin ^{ - 1}}\left( {\sin \left(-\frac{\pi}{3}\right)} \right)\)

Since \(-\frac{\pi}{2} \le -\frac{\pi}{3} \le \frac{\pi}{2}\), we can use the property \({\sin ^{ - 1}}(\sin \theta) = \theta\):

\({\sin ^{ - 1}}\left( {\sin \left(-\frac{\pi}{3}\right)} \right) = -\frac{\pi}{3}\)

So, the second part of the value of expression is \(-\frac{\pi}{3}\). Understanding how to handle inverse trigonometric functions for angles outside the principal range is key to correctly evaluate.

Calculating the Final Value of the Expression

Now we combine the results from the two parts to find the total value of the expression:

Value = \({\cos ^{ - 1}}\left( {\cos \frac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\sin \frac{{5\pi }}{3}} \right)\)

Value = \(\frac{\pi}{3} + \left(-\frac{\pi}{3}\right)\)

Value = \(\frac{\pi}{3} - \frac{\pi}{3}\)

Value = \(0\)

Thus, the final value of expression is 0. This problem helps reinforce concepts related to inverse trigonometric functions and trigonometric identities.