Understanding the Inverse Trigonometric Function Problem

The problem asks us to simplify the given inverse trigonometric function \(f(x) = {\sin ^{ - 1}}\left[ {\frac{{\sqrt 3 }}{2}x - \frac{1}{2}\sqrt {1 - {x^2}} } \right]\) for the specified domain \(x \in \left[ { - \frac{1}{2},1} \right]\). Simplifying such an expression typically involves using trigonometric identities and appropriate substitutions, keeping the domain restriction in mind.

Applying Trigonometric Substitution to Simplify the Expression

The expression inside the inverse sine function, \(\frac{{\sqrt 3 }}{2}x - \frac{1}{2}\sqrt {1 - {x^2}}\), involves \(x\) and \(\sqrt{1-x^2}\). This structure strongly suggests a trigonometric substitution. Let's try substituting \(x = \sin \theta\). Since \(x \in \left[ { - \frac{1}{2},1} \right]\), and the principal range of \({\sin ^{ - 1}}x\) is \(\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]\), the substitution \(x = \sin \theta\) implies \(\theta = {\sin ^{ - 1}}x\). For \(x \in \left[ { - \frac{1}{2},1} \right]\), the corresponding range for \(\theta\) is \(\left[ {\sin ^{ - 1}}\left( { - \frac{1}{2}} \right), {\sin ^{ - 1}}\left( 1 \right) \right]\), which is \(\left[ - \frac{\pi}{6}, \frac{\pi}{2} \right]\).

Now, substitute \(x = \sin \theta\) into the function \(f(x)\): \[f(x) = {\sin ^{ - 1}}\left[ {\frac{{\sqrt 3 }}{2}\sin \theta - \frac{1}{2}\sqrt {1 - \sin^2 \theta } } \right]\] Using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\): \[f(x) = {\sin ^{ - 1}}\left[ {\frac{{\sqrt 3 }}{2}\sin \theta - \frac{1}{2}\sqrt {\cos^2 \theta } } \right]\] Since \(\theta \in \left[ - \frac{\pi}{6}, \frac{\pi}{2} \right]\), \(\cos \theta \ge 0\). Therefore, \(\sqrt{\cos^2 \theta} = \cos \theta\). \[f(x) = {\sin ^{ - 1}}\left[ {\frac{{\sqrt 3 }}{2}\sin \theta - \frac{1}{2}\cos \theta } \right]\]

Using Trigonometric Identities for Simplification

The expression inside the inverse sine is \(\frac{{\sqrt 3 }}{2}\sin \theta - \frac{1}{2}\cos \theta\). This looks like the expansion of \(\sin(A-B)\) or \(\cos(A+B)\). Let's try to match it with \(\sin(A-B) = \sin A \cos B - \cos A \sin B\).

We can rewrite the expression as \(\sin \theta \cdot \left( \frac{\sqrt 3}{2} \right) - \cos \theta \cdot \left( \frac{1}{2} \right)\). We know that \(\cos \frac{\pi}{6} = \frac{\sqrt 3}{2}\) and \(\sin \frac{\pi}{6} = \frac{1}{2}\). Substituting these values:

\[ \sin \theta \cos \frac{\pi}{6} - \cos \theta \sin \frac{\pi}{6} \] This is the expansion of \(\sin\left(\theta - \frac{\pi}{6}\right)\). So, the function becomes: \[f(x) = {\sin ^{ - 1}}\left[ {\sin\left(\theta - \frac{\pi}{6}\right)} \right]\]

Considering the Domain Restriction and Final Simplification

For \({\sin ^{ - 1}}(\sin y)\) to simplify to \(y\), the argument \(y\) must be within the principal range of \({\sin ^{ - 1}}\), which is \(\left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]\). In our case, \(y = \theta - \frac{\pi}{6}\). We need to check if this argument is within the required range for all \(\theta \in \left[ - \frac{\pi}{6}, \frac{\pi}{2} \right]\).

Let's find the range of \(\theta - \frac{\pi}{6}\): Starting with \(\theta \in \left[ - \frac{\pi}{6}, \frac{\pi}{2} \right]\), subtract \(\frac{\pi}{6}\) from all parts: \[ - \frac{\pi}{6} - \frac{\pi}{6} \le \theta - \frac{\pi}{6} \le \frac{\pi}{2} - \frac{\pi}{6} \] \[ - \frac{2\pi}{6} \le \theta - \frac{\pi}{6} \le \frac{3\pi}{6} - \frac{\pi}{6} \] \[ - \frac{\pi}{3} \le \theta - \frac{\pi}{6} \le \frac{2\pi}{6} \] \[ - \frac{\pi}{3} \le \theta - \frac{\pi}{6} \le \frac{\pi}{3} \] The interval \(\left[ - \frac{\pi}{3}, \frac{\pi}{3} \right]\) is fully contained within the principal range of \({\sin ^{ - 1}}\), which is \(\left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]\).

Therefore, for the given domain, we can simplify \({\sin ^{ - 1}}\left[ {\sin\left(\theta - \frac{\pi}{6}\right)} \right]\) directly to \(\theta - \frac{\pi}{6}\).

Finally, substitute back \(\theta = {\sin ^{ - 1}}x\): \[f(x) = {\sin ^{ - 1}}x - \frac{\pi}{6}\] This is the simplified form of the given inverse trigonometric function for the domain \(x \in \left[ { - \frac{1}{2},1} \right]\). This demonstrates how to simplify an inverse trigonometric function by using appropriate trigonometric substitution and verifying the resulting argument's range.