Problem 6 - Calculus Single Variable

Question

Calculate the value of the given inverse trigonometric function at the given point. $$ \arccos (-\sqrt{3} / 2) $$

Step-by-Step Solution

Answer

\( \arccos(-\sqrt{3}/2) = 5\pi/6 \)

Step 1: Understanding the Problem

The exercise asks us to calculate the value of an inverse trigonometric function, specifically the inverse cosine or arccosine. The expression given is \( \arccos(-\sqrt{3}/2) \). This means we need to find the angle \( \theta \) such that \( \cos(\theta) = -\sqrt{3}/2 \).

Step 2: Finding Possible Angles

The cosine function achieves the value \(-\sqrt{3}/2\) at angles in the second and third quadrants. These correspond to the angles \( \theta = \pi - \pi/6 = 5\pi/6 \) (second quadrant) and \( \theta = \pi + \pi/6 = 7\pi/6 \) (third quadrant).

Step 3: Applying the Range of Arccos

The range of \( \arccos(x) \) (for \( -1 \leq x \leq 1 \)) is from \( 0 \) to \( \pi \). Therefore, the valid angle within this range is \( 5\pi/6 \), as \( 7\pi/6 \) is outside of this range.

Step 4: Concluding the Calculation

Since the angle \( 5\pi/6 \) is within the range of the \( \arccos \) function and satisfies \( \cos(5\pi/6) = -\sqrt{3}/2 \), we conclude that \( \arccos(-\sqrt{3}/2) = 5\pi/6 \).

Key Concepts

Arccos Function

The arccos function, or inverse cosine function, is fundamental in trigonometry. Its primary role is to determine an angle given the cosine of that angle. When we say \(\arccos(x)\), we are looking for the angle \(\theta \) so that \(\cos(\theta) = x\). The range of the arccos function is particularly important to understand. It provides outputs only between \(0\) and \(\pi\) (or 0 to 180 degrees), which ensures we always get a principal value for the angle.
Unlike the regular cosine function which can generate outputs beyond this range through different rotations on the unit circle, the arccos function keeps things simple by sticking to these conventional angles.

• The input value \(x\) must fall within \(-1\leq x \leq 1\). This corresponds to the valid range of cosine values.
• The output angle \(\theta\) will always be between \(0\) and \(\pi\).

Thus, whenever you encounter an arccos problem, always check both the input validity and the output range.

Angle Calculations

In angle calculations using the arccos function, the goal is to find an angle whose cosine corresponds to the input value. Taking our specific problem, \(\arccos(-\sqrt{3}/2)\), we first identify the angles whose cosine gives us \(-\sqrt{3}/2\). Cosine being an even function implies reflection symmetry across the y-axis in the cosine graph. In practical terms, this means:

• We look for angles in the second quadrant (where cosine is negative) such as \(\theta = \pi - \pi/6 = 5\pi/6\).
• We also consider third quadrant angles like \(\theta = \pi + \pi/6 = 7\pi/6\).

However, remember the range of the arccos function constrains the valid answer to only those angles in the second quadrant. Therefore, for \(\arccos(-\sqrt{3}/2)\), we conclude \(5\pi/6\) fits within this limited scope.

Trigonometry Concepts

Trigonometry is all about the relationships between the angles and sides of triangles, particularly right triangles. Understanding inverse trigonometric functions like arccos is essential for solving problems involving these relationships. Every trigonometric function has an inverse counterpart, allowing us to backtrack from ratios to angles.

When dealing with cosine, recall that it's one of the basic trigonometric ratios, defined as the adjacent side over the hypotenuse in a right triangle. The negative value of \(\cos(\theta)\) indicates that the angle is on the left-hand side of the unit circle, either in the second or third quadrant, where cosine is traditionally negative.

• Each trigonometric function and its inverse are interrelated and essential for calculating a vast array of geometric and analytical problems.
• Mastering these basic functions elevates our capability to handle complex scenarios involving angles and measurements.

With each trigonometric concept learned, solving exercises like \(\arccos(-\sqrt{3}/2)\) becomes a straightforward task of applying knowledge about function properties and angle characteristics.