Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

Draw the graph of \(y=\pi / 2-\arcsin x .\) Make a conjecture. Prove it.

Step-by-Step Solution

Answer

The graph of \(y = \frac{\pi}{2} - \arcsin(x)\) is the graph of \(\arccos(x)\).

Step 1: Identify the Original Function

The given function is in the form of \( y = \frac{\pi}{2} - \arcsin(x) \). Here, \( \arcsin(x) \) is the inverse sine function, which we need to understand before graphing.

Step 2: Recall the Domain and Range of \(\arcsin(x)\)

The function \( \arcsin(x) \) is defined for \( x \) in the domain \( [-1, 1] \) and takes values in the range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).

Step 3: Transform the Function

Transform \( y = \frac{\pi}{2} - \arcsin(x) \) by considering it as a vertical shift of the graph of \( y = -\arcsin(x) \), lifting it by \( \frac{\pi}{2} \). This reflects the graph across the x-axis and then moves it upward.

Step 4: Determine the Range

Since \( \arcsin(x) \) ranges from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \), \( y = \frac{\pi}{2} - \arcsin(x) \) will range from \( 0 \) to \( \pi \).

Step 5: Sketch the Graph

Plot the critical points: for \( x = -1 \), \( y = \pi \); for \( x = 0 \), \( y = \frac{\pi}{2} \); and for \( x = 1 \), \( y = 0 \). The graph will appear as an inverted arc that starts at the top and slopes downward as \( x \) moves from \(-1\) to \(1\).

Step 6: Formulate a Conjecture

The graph of \( y = \frac{\pi}{2} - \arcsin(x) \) suggests a complementary relationship between \( \arcsin(x) \) and the angle \( \frac{\pi}{2} \). This is similar to the identity \( \sin(\frac{\pi}{2} - \theta) = \cos(\theta) \).

Step 7: Prove the Conjecture

Given \( y = \frac{\pi}{2} - \arcsin(x) \), consider the identity \( y = \arccos(x) \). The identity \( \arccos(x) = \frac{\pi}{2} - \arcsin(x) \) is a trigonometric identity that holds as both sides measure the complementary angles in a right triangle.

Key Concepts

Arcsin

The function \( \arcsin(x) \) is the inverse of the sine function, but focuses only on a restricted part of it. Unlike the regular sine function, which is periodic and oscillates between -1 and 1, the \( \arcsin(x) \) function maps real numbers back to angles. This inverse function allows us to find the angle whose sine is a given number.

In the inverse world of trigonometry, we only consider the arcsin function for inputs between -1 and 1. This is important because sine values beyond this range are not defined for real angles.

Think about a right triangle: if you know the sine of an angle, you can use the arcsin function to find the actual angle. But, because angles can be different depending on their quadrant, \( \arcsin(x) \) returns values only from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \) radians, covering the first and fourth quadrants.

Domain and Range

In mathematics, especially with functions, the domain and range give us a complete picture of a function's behavior. Let's talk about \( \arcsin(x) \).

• **Domain**: This is essentially where the function can operate. For \( \arcsin(x) \), the domain is restricted to \([-1, 1]\). This means \( \arcsin(x) \) only "works" when \( x \) is within this interval.
• **Range**: This is where the outputs or results of the function lie. For \( \arcsin(x) \), the range is \([-\frac{\pi}{2}, \frac{\pi}{2}]\). These values are the angles that arcsin can produce as outputs.

Understanding these constraints is crucial when graphing functions because it sets the boundaries within which they exist and apply. In the exercise, knowing how \( \frac{\pi}{2} - \arcsin(x) \) transforms these outputs to values between \( 0 \) and \( \pi \) is key. It shows how the new function behaves differently from \( \arcsin(x) \) itself.

Graphing Functions

Graphing trigonometric functions is like painting a picture of their behavior. Let's explore how \( y = \frac{\pi}{2} - \arcsin(x) \) gets its shape.

When you graph \( \arcsin(x) \), you get a curve rising from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) as \( x \) moves from -1 to 1. The function \( y = \frac{\pi}{2} - \arcsin(x) \) inverts this graph. Imagine flipping \( \arcsin(x) \) over the x-axis and shifting it upward by \( \frac{\pi}{2} \). This produces what appears as an arch starting from the point \((x, y) = (-1, \pi)\) and ending at \((1, 0)\).

To draw this graph, remember these key steps:

• Identify critical points: For \( x = -1, 0, \) and \( 1 \), calculating \( y \) gives \( \pi, \frac{\pi}{2}, \) and \( 0 \) respectively.
• Sketch smoothly between these points: Connect them to show the continuous nature of the function.
• Remember the transformation: It’s crucial to understand that \( \frac{\pi}{2} - \arcsin(x) \) represents a reflection and a vertical shift of \( \arcsin(x) \).

Trigonometric Identities

Trigonometric identities are like shortcuts in understanding relationships between angles and their trigonometric functions. One important identity is \( \arccos(x) = \frac{\pi}{2} - \arcsin(x) \). This identity reveals how the complementary angles relate.

• **Complementary Angles**: In a right triangle, any two non-right angles add up to \( \frac{\pi}{2} \). Therefore, if one angle is \( \arcsin(x) \), the other must be \( \arccos(x) \). This reflects the identity \( \sin(\frac{\pi}{2} - \theta) = \cos(\theta) \).
• **Visual Understanding**: Visualizing these angles in a triangle helps cement the identity. Imagine a circle where each important angle provides a piece of this big trigonometric puzzle.

Utilizing identities like the ones we've explored simplifies complex trigonometric problems, allowing students to see the broader connection across different trigonometric relationships. This underpinning concept supports smooth transitions between different trigonometric functions and their inverses.