Problem 72 - Precalculus: Functions and Graphs

Question

Verify the Identity. $$\arccos x+\arccos \sqrt{1-x^{2}}=\frac{\pi}{2}, 0 \leq x \leq 1$$

Step-by-Step Solution

Answer

The identity is verified as true; \( \arccos(x) + \arccos(\sqrt{1-x^2}) = \frac{\pi}{2} \).

Step 1: Understand the Range and Function

We need to verify the identity \( \arccos(x) + \arccos(\sqrt{1-x^2}) = \frac{\pi}{2} \) under the range \( 0 \leq x \leq 1 \). The inverse cosine function, \( \arccos \), returns an angle whose cosine is the input value. The range of \( \arccos(x) \) is \([0, \pi]\) and since \( x \) is between 0 and 1, the angle will be between 0 and \( \frac{\pi}{2} \).

Step 2: Use Cosine Identity

Recall the identity \( \cos(\theta) = x \) and \( \cos(\pi/2 - \theta) = \sin(\theta) \). Given \( \arccos(x) = \theta \), this implies \( x = \cos(\theta) \) and \( \sqrt{1-x^2} = \sin(\theta) \). Thus, \( \arccos(\sqrt{1-x^2}) = \arcsin(x) \) since arcsine gives the angle whose sine is the input value.

Step 3: Sum of Angles Identity

We need to show that \( \arccos(x) + \arccos(\sqrt{1-x^2}) \) is equal to \( \frac{\pi}{2} \). Using \( \arccos(x) = \theta \), and knowing \( \arccos(\sqrt{1-x^2}) = \arcsin(x) \), we have \( \arccos(x) + \arcsin(x) = \frac{\pi}{2} \), which is a known trigonometric identity.

Step 4: Confirm the Identity

The identity \( \arccos(x) + \arcsin(x) = \frac{\pi}{2} \) naturally leads to \( \arccos(x) + \arccos(\sqrt{1-x^2}) = \frac{\pi}{2} \) by the properties of the inverse trigonometric functions. Hence, the given equation holds true for \( 0 \leq x \leq 1 \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions help us find angles when we know trigonometric ratios like sine, cosine, and tangent. They are crucial for solving equations in trigonometry. Inverse functions essentially "undo" the operations done by the basic trigonometric functions. For example, if we know that

\( \cos(\theta) = x \),

the inverse function \( \arccos(x) \) gives back the angle \( \theta \). This makes it invaluable for finding unknown angles in various applications, such as physics and engineering. It's important to remember that inverse trigonometric functions, by their nature, only return angles within specific ranges to ensure they are uniquely defined. These ranges, such as

\([0, \pi]\) for \( \arccos \) and \([-\frac{\pi}{2}, \frac{\pi}{2}]\) for \( \arcsin \),

help maintain this uniqueness. The relationship between cosine and its inverse is a fundamental concept that helps us solve equations by providing a pathway from ratios back to angle measures.

Arccos

The function \( \arccos(x) \) is termed the "arc cosine" or "inverse cosine." It returns the angle whose cosine is \( x \). The range of \( \arccos \) is quite specific: it gives outputs between 0 and \( \pi \) radians. This is because the cosine function itself maps the angles from this range onto all potential values from -1 to 1. Hence, \( \arccos(x) \) is defined for inputs between -1 and 1. In practical terms, when you use \( \arccos(x) \) for a value of \( x \) that falls within

the standard range of the cosine function,

the result is always an angle between 0 and \( \pi \). This ensures that for any real x within

\([-1,1],\)

the output is always a valid angle. \( \arccos \) is crucial in equations where you are solving for angles when given a cosine value, such as

in the identity provided \( \arccos(x) + \arccos(\sqrt{1-x^2}) = \frac{\pi}{2} \).

Understanding the behavior of \( \arccos \) helps us ensure that the result is always a precise measure of an angle, crucial in many areas of mathematical problem-solving.

Range of Functions

In mathematics, the range of a function is the set of all output values it can produce. For inverse trigonometric functions, defining the range is key to ensuring each function has only one output for every valid input. This one-to-one relationship is essential for the inverse function to actually "reverse" the original function. For example, the function \( \arccos(x) \) has a range

of [0, \( \pi \)],

indicating it can return any angle between these two values. Similarly, \( \arcsin(x) \)'s range is

\([-\frac{\pi}{2}, \frac{\pi}{2}]\),

which means it produces angles in this interval. These range specifications are chosen so that each function can cover all possible outputs of its corresponding trigonometric function. Understanding the range is also helpful for verifying identities. Take the given identity: \( \arccos(x) + \arccos(\sqrt{1-x^2}) = \frac{\pi}{2} \). Knowing that the angles produced by each function do not exceed their established ranges assures us that the identity holds true. Thus, when working with trigonometric identities and transformations, keeping track of range is crucial for accurate and realistic solutions.