Question

If \( 0 \leq \alpha \leq \frac{\pi}{2} \) and \(\sin \left(\alpha - \frac{\pi}{12}\right) = \frac{1}{2}\), then \(\alpha\) is equal to:

• A. \(\frac{\pi}{6}\)
• B. \(\frac{\pi}{4}\)
• C. \(\frac{\pi}{3}\)
• D. \(\frac{5\pi}{12}\)
• E. \(\frac{7\pi}{12}\)

Hint:

When solving equations involving trigonometric functions, remember to consider the domain of the variable and adjust your solution to fall within this range.

Answer 1

The Correct Option is B

The equation \(\sin \left(\alpha - \frac{\pi}{12}\right) = \frac{1}{2}\) suggests that \(\alpha - \frac{\pi}{12}\) must equal angles where the sine is \(\frac{1}{2}\). 
These angles are typically \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\), but since \(\alpha\) is between \(0\) and \(\frac{\pi}{2}\), the angle \(\frac{5\pi}{6}\) can be disregarded.
Thus, set the equation to: \[ \alpha - \frac{\pi}{12} = \frac{\pi}{6} \] Solve for \(\alpha\): \[ \alpha = \frac{\pi}{6} + \frac{\pi}{12} = \frac{2\pi}{12} + \frac{\pi}{12} = \frac{3\pi}{12} = \frac{\pi}{4} \]