Problem 18 - Precalculus Mathematics for Calculus

Question

Use an appropriate Half-Angle Formula to find the exact value of the expression. $$\tan 15^{\circ}$$

Step-by-Step Solution

Answer

The exact value of \( \tan 15^{\circ} \) is \( 2 - \sqrt{3} \).

Step 1: Identify the Half-Angle Formula for Tangent

The half-angle formula for tangent is \( \tan \left(\frac{\theta}{2}\right) = \frac{1 - \cos \theta}{\sin \theta} \). In this case, we know \( \theta = 30^{\circ} \) because \( 15^{\circ} = \frac{30^{\circ}}{2} \).

Step 2: Calculate \( \cos 30^{\circ} \) and \( \sin 30^{\circ} \)

The value of \( \cos 30^{\circ} \) is \( \frac{\sqrt{3}}{2} \) and \( \sin 30^{\circ} \) is \( \frac{1}{2} \). You can obtain these values from the known trigonometric identities for special angles.

Step 3: Substitute into the Formula

Substitute \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \) and \( \sin 30^{\circ} = \frac{1}{2} \) into the half-angle formula: \( \tan 15^{\circ} = \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}} \).

Step 4: Simplify the Expression

Perform the subtraction in the numerator: \( 1 - \frac{\sqrt{3}}{2} = \frac{2}{2} - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2} \). Then divide by \( \frac{1}{2} \): \( \frac{\frac{2 - \sqrt{3}}{2}}{\frac{1}{2}} = 2 - \sqrt{3} \).

Step 5: Finalize the Exact Value

The exact value for \( \tan 15^{\circ} \) using the half-angle formula is \( 2 - \sqrt{3} \).

Key Concepts

Tangent

The tangent of an angle is a fundamental trigonometric function that can be thought of as the ratio of the opposite side to the adjacent side in a right-angled triangle. In trigonometry, it is often denoted as \( \tan \theta \). For instance, if you have an angle \( \theta \) in a right triangle:

• The side opposite the angle is the one facing it.
• The adjacent side is the one next to it.

This leads to the expression: \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \).
Another way to understand the tangent function is in terms of the unit circle. Here, it represents the slope of the line created by the angle \( \theta \) from the origin. Especially in terms of half-angle, like \( 15^{\circ} \) in this case, the half-angle formulas assist in calculating this ratio when traditional methods become complex, thus yielding exact values such as the exercise reveals.

Trigonometric Identities

Trigonometric identities are equations that hold true for all values of the involved angles. They are tools that simplify expressions and calculations involving trigonometric functions.
Some fundamental identities you might come across include:

• Pythagorean identities like \( \sin^2 \theta + \cos^2 \theta = 1 \).
• Reciprocal identities such as \( \tan \theta = \frac{1}{\cot \theta} \).

Understanding these identities catapults you into complex problem-solving, as they serve as blueprints for solving trigonometric equations.
In the context of the half-angle formula used in the exercise, knowing identities like the sine and cosine of \( 30^{\circ} \) simplifies finding the tangent of \( 15^{\circ} \). By utilizing the identity \( \tan \left( \frac{\theta}{2} \right) = \frac{1-\cos \theta}{\sin \theta} \), we can break down the problem to more manageable pieces using known values.

Special Angles

Special angles are those whose trigonometric values are known and often memorized because they appear frequently. Angles such as \( 30^{\circ}, 45^{\circ}, \) and \( 60^{\circ} \) are classic examples.
They are deemed 'special' because their sine, cosine, and tangent values can be expressed as simple fractions or radicals, making them convenient for calculations.

• For example, \( \sin 30^{\circ} = \frac{1}{2} \) and \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \).
• \( \tan 60^{\circ} \) is one since it simplifies to \( \sqrt{3} \).

In the given exercise of calculating \( \tan 15^{\circ} \) using the half-angle formula, you can easily substitute these known values into the formula.
This substitution provides a clear, exact answer rather than an approximation, which is critical in various fields like engineering or physics, where precision is crucial.