Problem 23 - Precalculus Mathematics for Calculus

Question

Use an appropriate Half-Angle Formula to find the exact value of the expression. $$\tan \frac{\pi}{8}$$

Step-by-Step Solution

Answer

\( \tan \frac{\pi}{8} = \sqrt{3 - 2\sqrt{2}} \)

Step 1: Identify the Half-Angle Formula for Tangent

To find \( \tan \frac{\pi}{8} \), we use the half-angle formula for tangent: \( \tan \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \). Here, we set \( \theta = \frac{\pi}{4} \) because \( \frac{\pi}{8} = \frac{\pi}{4} \times \frac{1}{2} \).

Step 2: Determine the Cosine of \( \theta \)

Find \( \cos \frac{\pi}{4} \). We know that \( \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \). This will be used in the half-angle formula for tangent.

Step 3: Substitute \( \cos \frac{\pi}{4} \) into Half-Angle Formula

Plug \( \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \) into the formula for \( \tan \frac{\theta}{2} \): \[ \tan \frac{\pi}{8} = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}}} \]

Step 4: Simplify the Expression

First, simplify the numerator: \[ 1 - \frac{\sqrt{2}}{2} = \frac{2}{2} - \frac{\sqrt{2}}{2} = \frac{2-\sqrt{2}}{2} \] Then, simplify the denominator:\[ 1 + \frac{\sqrt{2}}{2} = \frac{2}{2} + \frac{\sqrt{2}}{2} = \frac{2+\sqrt{2}}{2} \] Now substitute back:\[ \tan \frac{\pi}{8} = \sqrt{\frac{\frac{2-\sqrt{2}}{2}}{\frac{2+\sqrt{2}}{2}}} = \sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}} \]

Step 5: Rationalize the Expression

To eliminate the square root in the denominator, multiply both the numerator and the denominator by the conjugate of the denominator:\[ \frac{2-\sqrt{2}}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{(2-\sqrt{2})^2}{(2+\sqrt{2})(2-\sqrt{2})} \]After expanding and simplifying:Numerator: \[ (2-\sqrt{2})^2 = 4 - 4\sqrt{2} + 2 = 6 - 4\sqrt{2} \]Denominator: \[ (2+\sqrt{2})(2-\sqrt{2}) = 4 - 2 = 2 \]Thus:\[ \tan \frac{\pi}{8} = \sqrt{\frac{6 - 4\sqrt{2}}{2}} \]

Step 6: Complete Simplification

This can be further simplified by dividing both terms in the numerator by 2:\[ \frac{6 - 4\sqrt{2}}{2} = 3 - 2\sqrt{2} \]So:\[ \tan \frac{\pi}{8} = \sqrt{3 - 2\sqrt{2}} \]

Key Concepts

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions and hold true for all values of the variable where the functions are defined. These identities are powerful tools for simplifying expressions and solving trigonometric equations.

The tangent half-angle formula is one such identity and is particularly useful for finding precise trigonometric values of angles, such as \( \tan \frac{\pi}{8} \). The formula states:

• \( \tan \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \)

This formula allows us to express the tangent of a half-angle in terms of the cosine of the angle. It's essential whenever angles are not easily found on the unit circle.

By setting \( \theta = \frac{\pi}{4} \) for our exercise, we take advantage of this identity and find that \( \frac{\pi}{8} \) is half of \( \frac{\pi}{4} \), which is a standard angle with a well-known cosine value.

Exact Trigonometric Values

Exact trigonometric values refer to the precise values of trigonometric functions at common angles. Memorizing these can greatly simplify problem-solving. For angles such as \( 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2} \), these values are typically found on the unit circle.

In this particular problem, we employ \( \cos \frac{\pi}{4} \), which is well known to be \( \frac{\sqrt{2}}{2} \). Having this exact value allows us to use the tangent half-angle formula effectively.

• Knowing such values helps replace numerical approximations with precise calculations.
• This avoids errors associated with rounding or measurement.

Using exact values is crucial in mathematics to ensure precise, accurate results in solutions, especially when dealing with irrational numbers.

Rationalization Techniques

Rationalization techniques involve removing radicals from the denominator of a fraction, resulting in a more simplified expression. This is often useful in making calculations more tractable or in achieving the exact form of a result.

In this exercise, we are given \( \tan \frac{\pi}{8} = \sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}} \). To simplify further, we multiply the numerator and denominator by the conjugate of the denominator \( 2-\sqrt{2} \). This process is an example of a standard rationalization technique:

• Numerator: \( (2-\sqrt{2})^2 = 4 - 4\sqrt{2} + 2 = 6 - 4\sqrt{2} \)
• Denominator: \( (2+\sqrt{2})(2-\sqrt{2}) = 4 - 2 = 2 \)

By rationalizing, the square root is now only in the numerator, leading to a cleaner, simplified solution: \( \tan \frac{\pi}{8} = \sqrt{3 - 2\sqrt{2}} \). This technique is widely applicable for trigonometric functions and algebraic manipulation.