Problem 26 - Precalculus Mathematics for Calculus

Question

Use an appropriate Half-Angle Formula to find the exact value of the expression. $$\tan \frac{5 \pi}{12}$$

Step-by-Step Solution

Answer

The exact value of \( \tan \frac{5\pi}{12} \) is \( 7 + 4\sqrt{3} \).

Step 1: Identify the given angle

The angle given is \( \frac{5\pi}{12} \). Convert it to degrees for easier reference: \( \frac{5\pi}{12} \times \frac{180}{\pi} = 75^\circ \).

Step 2: Recognize the need for a half-angle formula

Since \( 75^\circ \) is not a standard angle, we find relationships using half-angle formulas. Specifically, we recognize that \( 75^\circ \) is half of \( 150^\circ \).

Step 3: Set up the necessary half-angle formula

Use the half-angle formula for tangent: \( \tan \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \). Here, \( \theta = 150^\circ \).

Step 4: Calculate \( \cos 150^\circ \)

\( \cos 150^\circ = \cos(180^\circ - 30^\circ) = -\cos 30^\circ = -\frac{\sqrt{3}}{2} \).

Step 5: Apply the half-angle formula

Substitute \( \cos 150^\circ = -\frac{\sqrt{3}}{2} \) into the half-angle formula for tangent:\[ \tan 75^\circ = \sqrt{\frac{1 - (-\frac{\sqrt{3}}{2})}{1 + (-\frac{\sqrt{3}}{2})}} \]

Step 6: Simplify the expression

This simplifies to\[ \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}}} \]Further simplify:\[ \sqrt{\frac{\frac{2+\sqrt{3}}{2}}{\frac{2-\sqrt{3}}{2}}} = \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}} \]

Step 7: Rationalize and simplify further

Multiply numerator and denominator by conjugate to rationalize:\[ \frac{2+\sqrt{3}}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}} = \frac{4 + 2\sqrt{3} + 2\sqrt{3} + 3}{4 - 3} = 7 + 4\sqrt{3} \]

Key Concepts

Trigonometric Functions

Trigonometric functions are fundamental in mathematics. They're based on the angles of triangles, especially right triangles. These functions include sine (\(\sin\theta\)), cosine (\(\cos\theta\)), and tangent (\(\tan\theta\)).
Each function relates the angles with the sides of a triangle, helping to solve problems involving angles and their measurements.

• Sine (\(\sin\)) gives the ratio of the opposite side to the hypotenuse.
• Cosine (\(\cos\)) provides the ratio of the adjacent side to the hypotenuse.
• Tangent (\(\tan\)) finds the ratio of the opposite side to the adjacent side.

Using these functions, we can understand more about the shapes and relations in trigonometry, essential for navigation, physics, engineering, and more.

Angle Conversion

Converting angles from one form to another is a necessary skill in trigonometry. Angles are usually measured in degrees or radians. Each representation has its use, with radians being favored in higher-level mathematics and degrees in basic trigonometry.
To convert between these, you use the formula: \[ 1\, radian = \frac{180}{\pi} \quad degrees \]Here's how it works:

• From radians to degrees: multiply the radian measure by \(\frac{180}{\pi}\).
• From degrees to radians: multiply the degree measure by \(\frac{\pi}{180}\).

For instance, the angle \(\frac{5\pi}{12}\) can be converted to degrees as \(75^\circ\), which is easier for some calculations.

Tan Function

The tangent function, one of the core trigonometric functions, is important for calculating angles and distances. Defined as:\[\tan\theta = \frac{\sin\theta}{\cos\theta}\]It provides the ratio of the length of the opposite side to the adjacent side in a right triangle. This function varies based on the angle, having distinct properties:

• Tangent has periodicity of \(180^\circ\) or \(\pi\) radians, meaning its value repeats every \(\pi\) radians.
• It's undefined where cosine equals zero, typically at angles of \(90^\circ + n\cdot180^\circ\).

Using the half-angle formula, you can derive tangent values for non-standard angles. The half-angle formula for tangent \(\tan \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}\)helps in finding these values, as seen with \(75^\circ\).

Exact Values

Exact values in trigonometry allow us to calculate precise trigonometric functions without a calculator's rounded decimals. For angles like \(0^\circ, 30^\circ, 45^\circ, 60^\circ, \text{and} 90^\circ\), we have these exact values memorized.
When dealing with angles like \(75^\circ\), we use identities or transformations like the half-angle formula to derive them.
Here's the nature of exact values:

• Exact values are crucial in mathematical proofs and solving equations precisely.
• For compound or non-standard angles, derive by decomposition into known angles or use formulas.
• This practice helps when calculators are unavailable, like in written exams or theoretical work.

Thus, the utilization of the half-angle formula showcases how mathematics allows us to transform known values into exact solutions for various angles.