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Chapter 3: Problem 62

Determine whether the statement is true or false. Explain your answer. We can conclude from the derivatives of \(\sin ^{-1} x\) and \(\cos ^{-1} x\) that \(\sin ^{-1} x+\cos ^{-1} x\) is constant.

Short Answer

Expert verified
True, because \\( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \\) always.

Step by step solution

01

Understand the Statement

The statement tries to conclude whether the sum of the inverse sine function \( \sin^{-1} x \) and the inverse cosine function \( \cos^{-1} x \) results in a constant value based on their derivatives.
02

Recall Inverse Trigonometric Identities

Recall that both \( \sin^{-1} x \) and \( \cos^{-1} x \) are inverse trigonometric functions and that \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \) for \[ -1 \leq x \leq 1. \] So, \( \sin^{-1} x + \cos^{-1} x \) is a constant equal to \( \frac{\pi}{2}\).
03

Differentiate the Functions

To check if a function is constant, take its derivative: \( \frac{d}{dx}(\sin^{-1} x + \cos^{-1} x) \). The derivative of \( \sin^{-1} x \) is \( \frac{1}{\sqrt{1-x^2}} \) and the derivative of \( \cos^{-1} x \) is \( -\frac{1}{\sqrt{1-x^2}} \).
04

Add the Derivatives

Combine the derivatives: \[ \frac{d}{dx}(\sin^{-1} x + \cos^{-1} x) = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} = 0. \] Because the derivative is zero, \( \sin^{-1} x + \cos^{-1} x \) is indeed a constant function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse Sine Function
The inverse sine function, commonly denoted as \( \sin^{-1} x \) or arcsin\( x \), is a function that reverses the sine function. It is used to find the angle whose sine is \( x \). For example, if \( \sin \theta = x \), then \( \sin^{-1} x = \theta \).
This function has a range of \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \) and a domain of \( [-1, 1] \). This means that \( \sin^{-1} x \) will only give you angles within this interval.
Here are some key points to remember about the inverse sine function:
  • The derivative of \( \sin^{-1} x \) is \( \frac{1}{\sqrt{1-x^2}} \), representing how the function changes at different values of \( x \).
  • The function is odd, meaning that \( \sin^{-1}(-x) = -\sin^{-1}(x) \).
Inverse Cosine Function
The inverse cosine function is denoted as \( \cos^{-1} x \) or arccos\( x \). It helps to find the angle whose cosine is \( x \). In terms of angles, if \( \cos \theta = x \), then \( \cos^{-1} x = \theta \).
The range of \( \cos^{-1} x \) is \( [0, \pi] \) and the domain is \( [-1, 1] \). This means you will always get an angle in radians between \( 0 \) and \( \pi \).
Key details about the inverse cosine function include:
  • The derivative of \( \cos^{-1} x \) is \( -\frac{1}{\sqrt{1-x^2}} \). This negative sign indicates its inverse behavior compared to \( \sin^{-1} x \).
  • Similar to inverse sine, \( \cos^{-1} x \) is neither an even nor odd function.
Trigonometric Identities
Trigonometric identities are fundamental relationships between different trigonometric functions. They simplify expressions and make problem-solving more straightforward. One of the most important trigonometric identities involving inverse trigonometric functions is:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
This identity is valid for \( -1 \leq x \leq 1 \) and implies that the sum of the inverse sine and inverse cosine of the same angle \( x \) is always \( \frac{\pi}{2} \).
Here's how this identity helps:
  • It confirms that \( \sin^{-1} x + \cos^{-1} x \) is a constant value, regardless of \( x \) in the specified range.
  • This can be verified by calculating the derivative of the sum, which gives zero, implying a constant function.
  • It provides an intriguing glimpse into the symmetry of the unit circle, where sine and cosine complement each other's angles to make a right angle (\( \frac{\pi}{2} \)).

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