Question

Draw the graphs of $$ y=\arcsin x \quad \text { and } \quad y=\arctan \left(x / \sqrt{1-x^{2}}\right) $$ using the same axes. Make a conjecture. Prove it.

Short answer

The graphs are identical for \(-1 < x < 1\), proving the functions are equal in this interval.

Step-by-step solution

Understand the functions

We have two functions to analyze: 1. \( y = \arcsin x \), which is the inverse of the sine function. It is defined for \( -1 \leq x \leq 1 \) and outputs values in the range \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \).2. \( y = \arctan \left(\frac{x}{\sqrt{1-x^2}}\right) \), which involves both sine and tangent functions. This expression is valid for \( -1 < x < 1 \) as the denominator cannot be zero.

Graph the function \( y = \arcsin x \)

To graph \( y = \arcsin x \), plot points for \( x \) values between -1 and 1. Since \( \arcsin x \) is the inverse function of the sine, these functions will swap the roles of inputs and outputs compared to the standard sine graph.

Simplify and graph \( y = \arctan \left(\frac{x}{\sqrt{1-x^2}}\right) \)

Observe that when \( x = \sin \theta \), the expression \( \frac{x}{\sqrt{1-x^2}} = \tan \theta \). Thus, \( y = \arctan \tan \theta = \theta \), implying \( y = \arcsin x \). Hence, this graph overlaps with \( y = \arcsin x \) for \( -1 < x < 1 \).

Conjecture

The graphs of \( y = \arcsin x \) and \( y = \arctan \left(\frac{x}{\sqrt{1-x^2}}\right) \) appear to be identical within the interval \( -1 < x < 1 \). Thus, the conjecture is that these two expressions are equivalent over their domain.

Prove the conjecture

By letting \( x = \sin \theta \), we have \( \frac{x}{\sqrt{1-x^2}} = \tan \theta \). Therefore, \( \arctan \tan \theta = \theta \), which is \( \arcsin x \). This demonstrates that for \( -1 < x < 1 \), \( \arcsin x = \arctan \left(\frac{x}{\sqrt{1-x^2}}\right) \), confirming the conjecture.

Key concepts

Arcsin Function

The arcsin function, or inverse sine function, is an essential concept in trigonometry. It reverses the sine operation, allowing us to find an angle whose sine value is a given number. This function is denoted as \( y = \arcsin x \) and is only defined for input values \( x \) between -1 and 1.
The output of \( \arcsin x \) is an angle, measuring in radians, that lies within the range \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \). This ensures that any output is a valid angle on the unit circle.
Key points about the arcsin function include:

• Domain: \( -1 \leq x \leq 1 \)
• Range: \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \)
• Graph: It has an S-shape that passes through the origin.
• Swap roles: Since arcsin is the inverse of sine, it swaps the roles of inputs and outputs compared to the sine graph.

Understanding the arcsin function helps when comparing it to transforms or similar trigonometric expressions like \( \arctan \).

Arctan Function

The arctan function, or inverse tangent function, finds the angle whose tangent is a specific number. It is typically expressed as \( y = \arctan x \), which takes any real number and outputs an angle.
The angle produced by \( \arctan x \) is within the range of \( -\frac{\pi}{2} < y < \frac{\pi}{2} \). It is crucial in converting ratios of sides in rectangular triangles back to angles.
Key features of the arctan function include:

• Domain: All real numbers
• Range: \( -\frac{\pi}{2} < y < \frac{\pi}{2} \)
• Graph: It smoothly approaches the horizontal asymptotes at \( y = \pm\frac{\pi}{2} \).
• Application: Used in resolving angles in trigonometric identities.

The exercise dealt with the expression \( y = \arctan \left(\frac{x}{\sqrt{1-x^2}}\right) \), where an understanding of the function's properties supports the analysis of the function's behavior in relation to \( \arcsin x \).

Trigonometric Identities

Trigonometric identities are equations that express relationships among the trigonometric functions. These identities are fundamental when simplifying complex trigonometric expressions and equations.
Some crucial trigonometric identities that come in handy include:

• Pythagorean Identity: \( \sin^2 \theta + \cos^2 \theta = 1 \)
• Tangent Identity: \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)
• Inverse Identities: \( \arcsin(\sin \theta) = \theta \) for \( \theta \) in the range \( -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \), among others.

In the given problem, noticing that \( x = \sin \theta \) simplifies \( \frac{x}{\sqrt{1-x^2}} \) to \( \tan \theta \) enables the conjecture that \( \arcsin x = \arctan \left(\frac{x}{\sqrt{1-x^2}}\right) \).
Utilizing identities allows us to bridge the gap between different trigonometric expressions, simplifying them effectively for analysis or computational use.