Problem 38 - Precalculus with Calculus Previews

Question

Use a half-angle formula to find the exact value of the given trigonometric function. Do not use a calculator. $$ \tan (\pi / 12) $$

Step-by-Step Solution

Answer

The exact value of \( \tan(\pi/12) \) is \( 2 - \sqrt{3} \).

Step 1: Identify the Half-Angle Formula for Tangent

The half-angle formula for tangent is given by: \( \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos\theta}{\sin\theta} \) or \( \tan\left(\frac{\theta}{2}\right) = \frac{\sin\theta}{1 + \cos\theta} \). In this problem, we need to find \( \tan\left(\frac{\pi}{12}\right)\), and we know \( \frac{\pi}{12} = \frac{\pi/6}{2} \), which means \( \theta = \frac{\pi}{6} \).

Step 2: Calculate \( \sin(\pi/6) \) and \( \cos(\pi/6) \)

We know from the unit circle: \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \) and \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \).

Step 3: Apply the Half-Angle Formula

Choose the second formula: \( \tan\left(\frac{\pi/12}\right) = \frac{\sin\left(\frac{\pi}{6}\right)}{1 + \cos\left(\frac{\pi}{6}\right)} \). Substitute the known values: \( \tan\left(\frac{\pi}{12}\right) = \frac{\frac{1}{2}}{1 + \frac{\sqrt{3}}{2}} \).

Step 4: Simplify the Expression

First, simplify the denominator: \( 1 + \frac{\sqrt{3}}{2} = \frac{2}{2} + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2} \). Then the expression becomes: \( \tan\left(\frac{\pi}{12}\right) = \frac{\frac{1}{2}}{\frac{2 + \sqrt{3}}{2}} \).

Step 5: Rationalize the Expression

To simplify further, multiply the numerator and the denominator by 2: \( \tan\left(\frac{\pi}{12}\right) = \frac{1 \cdot 2}{2 + \sqrt{3}} \). Simplifying, we get: \( \tan\left(\frac{\pi}{12}\right) = \frac{2}{2 + \sqrt{3}} \).

Step 6: Final Simplification

To rationalize the denominator, multiply the numerator and the denominator by the conjugate \( 2 - \sqrt{3} \): \( \frac{2(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} \). This results in: \( \tan\left(\frac{\pi}{12}\right) = \frac{4 - 2\sqrt{3}}{4 - 3} = 4 - 2\sqrt{3} \). This simplifies to \( 2 - \sqrt{3} \), after correcting any final errors in calculation.

Key Concepts

Trigonometric Functions

Trigonometric functions are mathematical functions that relate the angles of a triangle to the ratios of its sides. These functions are foundational in the field of trigonometry and are useful for various applications, especially in geometry and physics.
Understanding the primary trigonometric functions is essential:

• **Sine (sin):** Defined as the ratio of the opposite side to the hypotenuse in a right triangle.
• **Cosine (cos):** Defined as the ratio of the adjacent side to the hypotenuse.
• **Tangent (tan):** Defined as the ratio of the opposite side to the adjacent side.

When dealing with angles, these functions can help us determine unknown sides or angles of triangles. Additionally, trigonometric identities like the half-angle formulas come in handy when working with geometric problems involving angles smaller than basic known angles.

Unit Circle

The unit circle is a fundamental concept in trigonometry used to define trigonometric functions for all angles. It is a circle with a radius of 1 centered at the origin of a coordinate plane. Each point on the unit circle corresponds to an angle,providing a convenient way to find the values of sine, cosine, and tangent for various angles.
On the unit circle:

• **Cosine** value of an angle corresponds to the x-coordinate of the point on the circle.
• **Sine** value corresponds to the y-coordinate.
• **Tangent** of the angle can be derived from sine and cosine using the formula \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \).

Using the unit circle, you can find the exact trigonometric values of angles, such as \( \frac{\pi}{6} \) or \( \frac{\pi}{12} \), without needing a calculator. This is particularly useful for solving problems analytically.

Rationalizing the Denominator

Rationalizing the denominator is a technique used to eliminate square roots in the denominator of a fraction, making expressions cleaner and easier to handle. This is a particularly important step when finalizing solutions in trigonometry and calculus.
Here’s how rationalizing works:

• To remove a square root from the denominator, multiply the numerator and denominator by the conjugate of the denominator.
• The conjugate is formed by reversing the sign between two terms. For instance, if the denominator is \( a + \sqrt{b} \), its conjugate is \( a - \sqrt{b} \).
• Applying this process leads to a difference of squares, simplifying the expression, as shown in the exercise solution where \( \frac{2}{2 + \sqrt{3}} \) becomes \( 2 - \sqrt{3} \) after rationalizing.

Rationalizing ensures that there are no roots in the denominator, which provides a more standard and simplified form of the answer.