Problem 12 - Precalculus with Calculus Previews

Question

Use a sum or difference formula to find the exact value of the given trigonometric function. Do not use a calculator. $$ \tan \frac{7 \pi}{12} $$

Step-by-Step Solution

Answer

The exact value of \( \tan\frac{7\pi}{12} \) is \(-2-\sqrt{3}\).

Step 1: Identify the Angles

We need to express \( \frac{7\pi}{12} \) as a sum or difference of known angles. Notice that \( \frac{7\pi}{12} = \frac{3\pi}{12} + \frac{4\pi}{12} = \frac{\pi}{4} + \frac{\pi}{3} \).

Step 2: Recognize the Applicable Formula

For the angle \( \frac{7\pi}{12} = \frac{\pi}{4} + \frac{\pi}{3} \), use the tangent sum formula: \[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \] where \( a = \frac{\pi}{4} \) and \( b = \frac{\pi}{3} \).

Step 3: Find the Tangents of Known Angles

Calculate the tangent values of the known angles: \( \tan \frac{\pi}{4} = 1 \) and \( \tan \frac{\pi}{3} = \sqrt{3} \).

Step 4: Apply the Sum Formula

Plug the values into the formula: \[ \tan \left( \frac{\pi}{4} + \frac{\pi}{3} \right) = \frac{1 + \sqrt{3}}{1 - 1 \cdot \sqrt{3}} = \frac{1 + \sqrt{3}}{1 - \sqrt{3}} \].

Step 5: Simplify the Expression

To simplify \( \frac{1+\sqrt{3}}{1-\sqrt{3}} \), multiply the numerator and the denominator by the conjugate of the denominator: \( 1+\sqrt{3} \), resulting in \[ \frac{(1+\sqrt{3})(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})} = \frac{1 + 2\sqrt{3} + 3}{1 + 3} = \frac{4 + 2\sqrt{3}}{-2} = -2-\sqrt{3} \].

Key Concepts

Sum and Difference Formulas

In trigonometry, sum and difference formulas are essential tools that help simplify complex angle calculations. They are particularly useful when calculating the trigonometric functions of angles not typically found on a unit circle but instead expressed as the sum or difference of two known angles. For example, if you have the angle \( \frac{7\pi}{12} \), it can be broken down into \( \frac{\pi}{4} + \frac{\pi}{3} \). This allows you to compute the trigonometric function using more accessible angles.

• The sum formula for tangent is \( \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \).
• The difference formula for tangent is \( \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \).

By utilizing these formulas, you can find the exact values of angles that are not straightforward to calculate. For instance, applying the sum formula breaks a complex angle down into parts involving easier computations.

Exact Value of Trigonometric Functions

Finding the exact value of trigonometric functions often involves simplifying complex expressions without using a calculator. This requires an understanding of common angles and their trigonometric ratios, particularly those found on the unit circle, like \( \pi/6, \pi/4, \pi/3 \), and so forth.

• \( \tan \frac{\pi}{4} = 1 \)
• \( \tan \frac{\pi}{3} = \sqrt{3} \)

Knowing these values allows you to use the sum and difference formulas to evaluate other angles. For example, in our case, \( \tan \frac{7\pi}{12} \) requires that we use the angles \( \frac{\pi}{4} \) and \( \frac{\pi}{3} \) to find an exact trigonometric expression. This process avoids approximation errors that can occur with numerical calculations and ensures precision in your result.

Tangent Function

The tangent function is one of the fundamental trigonometric functions, often expressed in terms of sine and cosine as \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). It is periodic with a period of \( \pi \), meaning its values repeat every \( \pi \) units.
In the context of the sum formula, the tangent function exhibits a unique combination property:

• When adding angles, \( \tan(a + b) \) does not simply sum the tangents; instead, it requires the formula \( \frac{\tan a + \tan b}{1 - \tan a \tan b} \). This accounts for interactions between the angles.
• Tangent reflects changes in the slope of the angle, with vertical asymptotes (undefined values) whenever \( \cos \theta = 0 \), such as at \( \frac{\pi}{2}, \frac{3\pi}{2} \), etc.

By understanding the behavior and properties of the tangent function, students can more effectively leverage it in various trigonometric applications such as solving equations, modeling periodic phenomena, and simplifying expressions.