Problem 25 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

The value of \(\cos \left(\frac{1}{2} \cos ^{-1}\left(\frac{1}{8}\right)\right)\) is equal to (a) \(\frac{3}{4}\) (b) \(-\frac{3}{4}\)

Step-by-Step Solution

Answer

The value of \(\cos \left(\frac{1}{2} \cos^{-1}\left(\frac{1}{8}\right)\right)\) is \(\frac{3}{4}\). Thus, the correct option is (a) \(\frac{3}{4}\).

Step 1: Determine the angle

First, identify the angle for which the cosine is \(\frac{1}{8}\). That means finding the angle \( \theta \) such that \(\cos (\theta) = \frac{1}{8}\). By the definition of inverse cosine (also known as arccosine), this angle is \( \theta = \cos^{-1}\left(\frac{1}{8}\right) \).

Step 2: Apply the half-angle formula

The half-angle formula for cosine is given by \(\cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{2}}\). In this case, \(\theta = \cos^{-1} \left(\frac{1}{8}\right)\), and we are looking for \(\cos \left(\frac{1}{2} \cos^{-1}\left(\frac{1}{8}\right)\right) = \cos \left(\frac{\theta}{2}\right)\). By substituting \(\theta \) into the half-angle formula, we obtain a value equal to \( \sqrt{\frac{1 + \frac{1}{8}}{2}} = \sqrt{\frac{9}{16}} \).

Step 3: Simplify the result

The square root of \(\frac{9}{16}\) is \(\frac{3}{4}\). Since \(\frac{\theta}{2}\) lies in the first quadrant where cosine is positive, the sign in front of the square root in the half-angle formula for cosine should be positive. So, the result is positive \(\frac{3}{4}\).

Key Concepts

Inverse Cosine Function

The inverse cosine function, often called arccosine, is a fundamental concept in trigonometry. It is denoted as \(\cos^{-1}(x)\). This function helps us find the angle whose cosine is a given number. For example, if \(\cos(\theta) = \frac{1}{8}\), the inverse cosine function provides us with \(\theta = \cos^{-1}\left(\frac{1}{8}\right)\).

A critical aspect of the inverse cosine function is its range. The values it returns lie within the interval \([0, \pi]\) radians, which corresponds to angles from \(0\) to \(180\) degrees. This restriction exists because the cosine function can map to the same value on different intervals. The defined range ensures that each input gives a unique result.

In applications, the inverse cosine function is used extensively, especially when working with angles in right triangles or when needing to determine angles from given cos values. Understanding this function is essential for solving problems related to angles and their measures.

Quadrant Analysis

Quadrant analysis in trigonometry helps us understand where an angle's terminal side is located on the unit circle. This is crucial for determining the signs of trigonometric functions like sine, cosine, and tangent. The unit circle is typically divided into four quadrants:

• 1st Quadrant: Both sine and cosine are positive.
• 2nd Quadrant: Sine is positive, cosine is negative.
• 3rd Quadrant: Both sine and cosine are negative.
• 4th Quadrant: Sine is negative, cosine is positive.

In the given exercise, \(\theta\) is found using the inverse cosine, \(\cos^{-1}\left(\frac{1}{8}\right)\), and lies in the 1st quadrant. This is because, by definition, the inverse cosine outputs an angle in the range \([0, \pi]\).

When applying the trigonometric half-angle identities, determining the correct quadrant is vital to assigning the correct sign to the result. For instance, when using the half-angle formula, the sign of the result depends on the quadrant in which the angle \(\frac{\theta}{2}\) falls.

Trigonometric Functions

Trigonometric functions are sine, cosine, and tangent. They relate the angles of a triangle to the lengths of its sides. These functions are foundational in mathematics, physics, and engineering, providing the basis for analyzing periodic phenomena.

Each trigonometric function has specific properties. For instance:

• The cosine function, \(\cos(\theta)\), returns the x-coordinate on the unit circle.
• The sine function, \(\sin(\theta)\), returns the y-coordinate on the unit circle.
• The tangent function, \(\tan(\theta)\), represents the slope of the line forming angle \(\theta\).

In the problem at hand, the use of the cosine function and its inverse plays a central role. By applying the half-angle identity, \(\cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{2}}\), we find the value of the cosine of the halved angle.

Understanding these functions' identities and transformations allows for problem-solving in contexts like oscillations, waves, and other cyclic phenomena.