Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>Given \(0 \leq x \leq \frac{1}{2}\), then the value of \(\sin^{-1}\left(\frac{x + \sqrt{1-x^2}}{2}\right) - \sin^{-1}x\) is</p>
<p>(a) \(1\)</p>
<p>(b) \(\sqrt{3}\)</p>
<p>(c) \(-1\)</p>
<p>(d) \(\frac{1}{3}\)</p>
Step-by-Step Solution
Key Concept: Use substitution with inverse sine and trigonometric identities to simplify the expression
<p>Simplify the expression using the substitution $x = \sin\theta$ where $\theta \in [0, \pi/6]$.</p>
Correct Answer: A