Question

Solve for $ \alpha $ if: $ \cos^{-1}(2 \sin \alpha) = \frac{47}{12} $

• A. \( \alpha = \sin^{-1} \left( \frac{47}{12} \right) \)
• B. \( \alpha = \cos^{-1} \left( \frac{47}{12} \right) \)
• C. \( \alpha = \sin^{-1} \left( \frac{2}{12} \right) \)
• D. None of these

Hint:

Whenever solving inverse trigonometric equations, always check the ranges of the functions. For instance, \( \cos^{-1}(x) \) is only valid for \( x \in [-1, 1] \), and similarly for \( \sin^{-1}(x) \).

Answer 1

The Correct Option is D

To solve for \( \alpha \) in the equation: \(\cos^{-1}(2 \sin \alpha) = \frac{47}{12}\)

we first need to acknowledge the domain restrictions of the inverse trigonometric functions. The range of \(\cos^{-1}(x)\) is \([0, \pi]\), and the valid domain for \(x\) within \(\cos^{-1}(x)\) is \([-1, 1]\). Thus, for the expression \(2 \sin \alpha\) to be valid, it must satisfy:

\[-1 \leq 2 \sin \alpha \leq 1\]

Dividing the entire inequality by 2 gives:

\[-\frac{1}{2} \leq \sin \alpha \leq \frac{1}{2}\]

Next, we examine the given equation:

\(\cos^{-1}(2 \sin \alpha) = \frac{47}{12}\)

This implies the angle \(\frac{47}{12}\) radians must correspond to a valid cosine value that 2 times the sine of \(\alpha\) can achieve. However, converting \(\frac{47}{12}\) radians to degrees or using common knowledge of trigonometric values shows that it falls outside typical ranges used for trigonometric tables, and must be evaluated correctly in radians.

The cosine of an angle outside the domain limits means the equation has no valid solution under real number constraints. Therefore, none of the offered options, each incorrectly implying valid output for \(\alpha\) based on given inputs, match possible outcomes.

Thus, the correct conclusion is: None of these