Question

Evaluate: \( \sin^{-1} \left( \sin \frac{3\pi}{5} \right) \).

Hint:

When evaluating inverse trigonometric functions, make sure the result lies within the principal range of the function. If necessary, use trigonometric identities to adjust the angle to the correct range.

Answer 1

To evaluate \( \sin^{-1} \left( \sin \frac{3\pi}{5} \right) \), we consider the definition of \( \sin^{-1}(x) \). The function \( \sin^{-1}(x) \) returns an angle \( \theta \) such that \( \sin(\theta) = x \) and \( \theta \) is in the interval \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \). Since \( \frac{3\pi}{5}>\frac{\pi}{2} \), the angle \( \frac{3\pi}{5} \) is outside the principal range of the inverse sine function. We must find an equivalent angle within \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \) that has the same sine value. Using the trigonometric identity \( \sin \left( \pi - x \right) = \sin x \), we can rewrite \( \sin \left( \frac{3\pi}{5} \right) \) as: \[\sin \left( \frac{3\pi}{5} \right) = \sin \left( \pi - \frac{3\pi}{5} \right) = \sin \left( \frac{5\pi - 3\pi}{5} \right) = \sin \left( \frac{2\pi}{5} \right).\] Now, the angle \( \frac{2\pi}{5} \) is within the required range \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \) because \( -\frac{\pi}{2} \le \frac{2\pi}{5} \le \frac{\pi}{2} \). Therefore, \[\sin^{-1} \left( \sin \frac{3\pi}{5} \right) = \sin^{-1} \left( \sin \frac{2\pi}{5} \right) = \frac{2\pi}{5}.\] The result is \( \boxed{\frac{2\pi}{5}} \).