Problem 18 - Calculus Single Variable

Question

Calculate the value of the given inverse trigonometric function at the given point. $$ \arccos (\cos (4 \pi / 3)) $$

Step-by-Step Solution

Answer

The value is \(\frac{2\pi}{3}\).

Step 1: Simplify Cosine Function

First, recognize that the input to the inverse cosine function \( hetha = \cos(\frac{4\pi}{3})\) needs to be simplified. The angle \(\frac{4\pi}{3}\) corresponds to an angle in the third quadrant on a unit circle. \(\cos(\theta)\) for \(\theta = \frac{4\pi}{3}\) is equal to \(-\frac{1}{2}\). Therefore, \(\cos(\frac{4\pi}{3}) = -\frac{1}{2}\).

Step 2: Apply the Inverse Cosine Function

Now substitute back into the inverse cosine function: \(\arccos(-\frac{1}{2})\). We need to find the angle \(x\) whose cosine is \(-\frac{1}{2}\). \(\arccos\) is defined for \(0\) to \(\pi\); hence the relevant angle is in this range.

Step 3: Determine the Principal Value of Arccos

The principal value of \(\arccos(-\frac{1}{2})\) within \(0\) to \(\pi\) is \(\frac{2\pi}{3}\). This is because cosine of \(\frac{2\pi}{3}\) is \(-\frac{1}{2}\).

Key Concepts

Cosine Function

The cosine function is a fundamental component of trigonometry, representing the ratio of the adjacent side to the hypotenuse of a right-angled triangle. Its notation, cos(θ), where θ is the angle, is used to find how much of a shadow an angle creates. For example, in our exercise, calculating \(\cos\left(\frac{4\pi}{3}\right)\) helps us understand the direction and length of the shadow relative to the angle \(\theta\).

The cosine function is periodic with a period of \(2\pi\), meaning every \(2\pi\) radians, it repeats its values. This periodic nature allows angles greater than \(2\pi\) or negative angles to be easily translated back to an existing value within a single cycle by adding or subtracting \(2\pi\) until the angle is within the desired range.

• Cosine is positive in the first and fourth quadrants.
• Negative in the second and third quadrants.

Understanding and identifying the specific value like \(-\frac{1}{2}\) in the correct quadrant is crucial, as shown when finding \(\cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2}\).

Unit Circle

The unit circle is a crucial concept in trigonometry. It is a circle in the coordinate plane with a radius of one unit centered at the origin (0,0). The unit circle allows us to easily calculate trigonometric functions such as sine, cosine, and tangent for different angles.

• The x-coordinate of a point on the unit circle represents the cosine of that angle.
• The y-coordinate represents the sine of that angle.

In the problem, \(\frac{4\pi}{3}\) radians is an angle located in the third quadrant of the unit circle. Here, the cosine value is negative due to the angle's location. This negative cosine value, \(-\frac{1}{2}\), can be identified on the unit circle by analyzing its x-coordinate.

Understanding where specific angles fall on the unit circle and how they correspond to high-school level trigonometry functions provides a visual and intuitive way to connect angles and their trigonometric values.

Principal Value

The principal value in inverse trigonometric functions is the most common reference angle for an angle's trigonometric function value, especially when the angle can have multiple solutions. Specifically, for the inverse cosine function, \(\arccos(\theta)\), the principal value is the angle within the range \(0\) to \(\pi\).

This is because, unlike cosine which can cycle through all four quadrants, the principal value definition restricts it to one, making it easier to predictably determine which angle corresponds to a given cosine output. In our exercise:

• Calculating \(\arccos(-\frac{1}{2})\), the angle \(\frac{2\pi}{3}\), lies in this principal value range because it is the standard angle for which the cosine value equals \(-\frac{1}{2}\).

This concept prevents ambiguity and ensures consistency, especially when dealing with trigonometric equations where such a value could refer to infinite angles. Hence, understanding and utilizing principal values is essential for solving inverse trigonometric problems correctly.