Problem 10 - 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. $$ \tan \frac{11 \pi}{12} $$

Step-by-Step Solution

Answer

The exact value is \(-2 + \sqrt{3}\).

Step 1: Identify the Angles for Sum or Difference

We need to express \( \frac{11 \pi}{12} \) as a sum or difference of angles for which the tangent values are known. Notice that \( \frac{11 \pi}{12} = \frac{3 \pi}{4} + \frac{\pi}{6} \). Both \( \frac{3 \pi}{4} \) and \( \frac{\pi}{6} \) have known tangent values.

Step 2: Apply the Sum Formula for Tangent

The sum formula for tangent is given by \( \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \). Here, let \( a = \frac{3 \pi}{4} \) and \( b = \frac{\pi}{6} \).

Step 3: Substitute Known Tangent Values

From trigonometric values, we know \( \tan \frac{3 \pi}{4} = -1 \) and \( \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \). Substitute these into the formula: \( \tan \left(\frac{3 \pi}{4} + \frac{\pi}{6}\right) = \frac{-1 + \frac{1}{\sqrt{3}}}{1 - (-1)\left(\frac{1}{\sqrt{3}}\right)} \).

Step 4: Simplify the Expression

Simplify the expression: \[ \tan \frac{11 \pi}{12} = \frac{-1 + \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{-\sqrt{3} + 1}{\sqrt{3} + 1} \]

Step 5: Rationalize the Denominator

Multiply the numerator and denominator by the conjugate of the denominator \( \sqrt{3} - 1 \):\[ = \frac{(-\sqrt{3} + 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{-3 + \sqrt{3} + \sqrt{3} - 1}{3 - 1} \]\[ = \frac{-4 + 2\sqrt{3}}{2} \]\[ = -2 + \sqrt{3} \]

Step 6: Final Expression

Hence the exact value of \( \tan \frac{11 \pi}{12} \) is \( -2 + \sqrt{3} \).

Key Concepts

Sum and Difference Formulas

The sum and difference formulas are essential tools to find the exact values of trigonometric functions, especially when angles are not directly available as part of the unit circle. The sum formula for tangent, for example, helps in breaking down complex angles into simpler, well-known ones. The formula is:

• \( \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \)

To use this formula effectively, you need to choose angles \(a\) and \(b\) such that their sum equals the original angle whose tangent is to be found. This was illustrated in the solution where \( \frac{11 \pi}{12} \) was expressed as \( \frac{3 \pi}{4} + \frac{\pi}{6} \), making it possible to apply the formula. By selecting these angles, we can substitute known tangent values to simplify the equation and solve for the exact trigonometric value.

Exact Trigonometric Values

Exact trigonometric values are specific numerical values that represent the trigonometric functions of standard angles. Angles such as \( \pi/6 \), \( \pi/4 \), and \( \pi/3 \) often have straightforward trigonometric values involving integers and simple fractions like \( \sqrt{2} \) or \( \sqrt{3} \). Knowing these values allows for quick substitution when solving expressions involving trigonometric identities.
For example, in the exercise, the tangent values for \( \frac{3\pi}{4} \) and \( \frac{\pi}{6} \) were used:

• \( \tan \frac{3\pi}{4} = -1 \)
• \( \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \)

These known values directly feed into the sum formula used to find the exact tangent of a non-standard angle.

Rationalization of Denominators

Rationalizing the denominator is a mathematical technique used to eliminate any irrational numbers, like square roots, from the bottom part of a fraction. This makes the expression look cleaner and easier to interpret.
When dealing with expressions in trigonometry, such as the tangent sum formula that introduces square roots in the denominators, rationalizing is often needed for simplification. In the example provided, rationalization involved:

• Multiplying both the numerator and the denominator by the conjugate of the denominator.
• The conjugate is essentially the denominator with its middle sign changed, i.e., \( \sqrt{3} - 1 \) for the denominator \( \sqrt{3} + 1 \).

This operation resulted in the fraction \( \frac{-3 + \sqrt{3} + \sqrt{3} - 1}{3 - 1} \) simplifying to \( -2 + \sqrt{3} \), which is a neatly expressed and exact value.