Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>If $x$ takes negative permissible value then $\sin^{-1} x$ is</p>
<p>(a) $\cos^{-1}(\sqrt{1-x^2})$</p>
<p>(b) $-\cos^{-1}(\sqrt{1-x^2})$</p>
<p>(c) $\cos^{-1}(\sqrt{x^2-1})$</p>
<p>(d) $\pi - \cos^{-1}(\sqrt{1-x^2})$</p>
Step-by-Step Solution
Key Concept: For negative $x$, $\sin^{-1} x$ is negative, and it relates to $\cos^{-1}(\sqrt{1-x^2})$ through the complementary angle relationship in the principal branches.
<p><strong>Step 1:</strong> For $x \in [-1, 0)$, we have $\sin^{-1} x \in \left[-\frac{\pi}{2}, 0\right)$ (negative values).</p><p><strong>Step 2:</strong> We use the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$, which gives $\sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x$.</p><p><strong>Step 3:</strong> If $\sin^{-1} x = \alpha$ where $\alpha < 0$, then $\sin \alpha = x$ and $\cos \alpha = \sqrt{1-x^2}$ (positive in the fourth quadrant).</p><p><strong>Step 4:</strong> Thus $\cos^{-1}(\sqrt{1-x^2}) = -\alpha = -\sin^{-1} x$, so $\sin^{-1} x = -\cos^{-1}(\sqrt{1-x^2})$.</p>
Correct Answer: B