91Ó°ÊÓ

The concept of \(\text{arcsin}\) might seem complex, but it's quite simple. It represents the inverse sine function. In mathematics, functions can have inverses, and the inverse of sine is the \(\text{arcsin}\), or \(\text{sin}^{-1}\). This function takes a value (usually a ratio that results from sine) and tells you the angle that produced that ratio.
For example, if \( \text{sin}(\theta) = x \), then \( \text{arcsin}(x) = \theta \). This means \(\theta\) is the angle whose sine value gives \(x\). Always remember, when dealing with \(\text{arcsin}\), the result will be an angle. This angle will always be in the range of \(-\frac{\text{Ï€}}{2} \text{ to } \frac{\text{Ï€}}{2} \), because that's where \( \text{sin}\) takes all its values.
Using our exercise, we set \( \theta = \text{arcsin}(x) \). From this, \( \text{sin}(\theta) = x \), giving us a base to substitute in other trigonometric identities.

One of the foundational principles in trigonometry is the Pythagorean Identity. Essentially, for any angle \( \theta \), the following relationship holds true: \(\text{sin}^2(\theta) + \text{cos}^2(\theta) = 1 \). This identity comes from the Pythagorean Theorem, applied to a right-angle triangle.
From the exercise, knowing \( \text{sin}(\theta) = x \), we substitute to get \(\text{cos}(\theta) = \text{sqrt}(1 - \text{sin}^2(\theta))\).
Substituting \( \text{sin}(\theta) = x \) gives us \( \text{cos}(\theta) = \text{sqrt}(1 - x^2) \). This step helps in seamlessly transitioning to our next calculation.

Trigonometric functions are at the heart of solving problems involving angles and triangles. They include sine (sin), cosine (cos), and tangent (tan), among others.
Each trigonometric function relates an angle to a ratio of sides in a right triangle. For example:
- \( \text{sin}(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \)
- \( \text{cos}(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \)
- \( \text{tan}(\theta) = \frac{\text{opposite}}{\text{adjacent}} \)

In this exercise, we need \( \text{tan}(\theta) \). We already know \( \text{sin}(\theta) = x \) and \( \text{cos}(\theta) = \text{sqrt}(1 - x^2) \). Therefore, \( \text{tan}(\theta) = \frac{\text{sin}(\theta)}{\text{cos}(\theta)} = \frac{x }{ \text{sqrt}(1 - x^2) } \). This gives us the equivalent algebraic expression for \( \text{tan}(\text{arcsin}(x)) \).