Problem 21 - Algebra and Trigonometry

Question

15–26 Use an appropriate half-angle formula to find the exact value of the expression. $$\tan \frac{\pi}{8}$$

Step-by-Step Solution

Answer

The exact value of \( \tan \frac{\pi}{8} \) is \( \sqrt{2} - 1 \).

Step 1: Identify the Problem

We are tasked to find the exact value for \( \tan \frac{\pi}{8} \) using a trigonometric identity. The expression represents a half-angle, which suggests using a half-angle formula.

Step 2: Recall Half-Angle Formula for Tangent

The half-angle identity for tangent is \( \tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta} \) or \( \tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta} \). We need to express \( \frac{\pi}{8} \) as a half-angle of another familiar angle.

Step 3: Express or Confirm Angle Relation

Note that \( \frac{\pi}{8} = \frac{\pi}{4} / 2 \). This means \( \frac{\pi}{8} = \frac{\theta}{2} \), where \( \theta = \frac{\pi}{4} \). Therefore, we can use \( \theta = \frac{\pi}{4} \) in the formula.

Step 4: Apply Half-Angle Formula

Using \( \tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta} \) with \( \theta = \frac{\pi}{4} \), we know \( \cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \). So, \( \tan \frac{\pi}{8} = \frac{1 - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} \).

Step 5: Simplify the Expression

Simplify the expression: \( \tan \frac{\pi}{8} = \frac{\frac{2}{2} - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{2 - \sqrt{2}}{\sqrt{2}} \). Multiply numerator and denominator by \( \sqrt{2} \) to rationalize the denominator. This results in \( \frac{2\sqrt{2} - 2}{2} \).

Step 6: Final Simplification

The final expression simplifies to \( \tan \frac{\pi}{8} = \sqrt{2} - 1 \). Therefore, the exact value is \( \tan \frac{\pi}{8} = \sqrt{2} - 1 \).

Key Concepts

trigonometric identities

Trigonometric identities are special equations that relate various trigonometric functions. They serve as essential tools in simplifying complex trigonometric expressions and solving trigonometry problems. Among these identities are the half-angle formulas, which provide a way to find the trigonometric functions of half of an angle.
For tangent, the half-angle formula can be expressed in two ways:

• \( \tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta} \) or
• \( \tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta} \)

These identities are particularly useful when we need to calculate the trigonometric value of angles that are not commonly found in trigonometry tables. By understanding and using these identities, we can find exact values efficiently.

exact values

To find the exact value of a trigonometric expression means to determine its true value using standard mathematical constants, rather than a decimal approximation. In many cases, this involves using known values of trigonometric functions for standard angles.
For the case of \( \tan \frac{\pi}{8} \), we utilize the half-angle formulas to express \( \frac{\pi}{8} \) in terms of \( \frac{\pi}{4} \), a standard angle with known trigonometric values:

• \( \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \)
• \( \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \)

Using these exact values, we achieve a simplified result without approximation: \( \tan \frac{\pi}{8} = \sqrt{2} - 1 \). This process of deriving exact values is crucial in mathematics, avoiding errors that might arise from rounding.

tangent

The tangent of an angle in a right-angled triangle is the ratio of the opposite side to the adjacent side. It is one of the fundamental trigonometric functions, alongside sine and cosine. For angles in the unit circle, tangent can also be expressed as the ratio of sine to cosine:

• \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)

In our exercise, we are tasked with finding \( \tan \frac{\pi}{8} \) using the half-angle formula. With the angle \( \frac{\pi}{8} \) effectively a half-angle of \( \frac{\pi}{4} \), we substituted into our formula to simplify and find:

• \( \tan \frac{\pi}{8} = \sqrt{2} - 1 \)

Understanding how tangent interplays with other trigonometric functions and identities helps to tackle complex trigonometry problems efficiently. This exercise illustrated a practical application of these principles to derive an exact value.