Problem 4 - Algebra 2 and Trigonometry

Question

In \(3-17,\) find the exact value of \(\tan (A+B)\) and of \(\tan (A-B)\) for each given pair of values. $$ A=45^{\circ}, B=60^{\circ} $$

Step-by-Step Solution

Answer

\( \tan(A+B) = -2 - \sqrt{3} \) and \( \tan(A-B) = -2 + \sqrt{3} \).

Step 1: Understand the Problem

We need to find the exact values of \( \tan(A+B) \) and \( \tan(A-B) \) for the given angles \( A = 45^{\circ} \) and \( B = 60^{\circ} \). This requires using the tangent addition and subtraction formulas.

Step 2: Recall Tangent Addition and Subtraction Formulas

The tangent addition formula is \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \). The tangent subtraction formula is \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \).

Step 3: Calculate \( \tan 45^{\circ} \) and \( \tan 60^{\circ} \)

Recall that \( \tan 45^{\circ} = 1 \) and \( \tan 60^{\circ} = \sqrt{3} \). These values will be used in the formulas.

Step 4: Apply the Tangent Addition Formula

Substitute \( \tan 45^{\circ} = 1 \) and \( \tan 60^{\circ} = \sqrt{3} \) into the tangent addition formula: \[ \tan(A + B) = \frac{1 + \sqrt{3}}{1 - 1\sqrt{3}} \] \[ \tan(A + B) = \frac{1 + \sqrt{3}}{1 - \sqrt{3}} \].

Step 5: Simplify \( \tan(A+B) \)

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

Step 6: Apply the Tangent Subtraction Formula

Substitute \( \tan 45^{\circ} = 1 \) and \( \tan 60^{\circ} = \sqrt{3} \) into the tangent subtraction formula: \[ \tan(A - B) = \frac{1 - \sqrt{3}}{1 + 1\sqrt{3}} \] \[ \tan(A - B) = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \].

Step 7: Simplify \( \tan(A-B) \)

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

Key Concepts

Tangent Addition Formula

The tangent addition formula is a fundamental trigonometric identity used to find the tangent of the sum of two angles:

• Formula: \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)

This formula comes in handy when you know the tangents of two angles, but you need to find the tangent of their sum.
It's particularly useful in various fields like physics, engineering, and architecture where angles often need precise calculations.
In our exercise, we are given two angles, \( A = 45^{\circ} \) and \( B = 60^{\circ} \), and asked to find \( \tan(A + B) \).
First, compute the tangent for each angle:

• \( \tan 45^{\circ} = 1 \)
• \( \tan 60^{\circ} = \sqrt{3} \)

Plug these values back into the formula:

• \[ \tan(A + B) = \frac{1 + \sqrt{3}}{1 - 1\sqrt{3}} = \frac{1 + \sqrt{3}}{1 - \sqrt{3}} \]

The calculation involves rationalizing the denominator, leading to the final simplified result.

Tangent Subtraction Formula

Just like the addition formula, the tangent subtraction formula helps us find the tangent of the difference between two angles:

• Formula: \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \)

This is a crucial identity, especially when dealing with inverse trigonometric functions or problems related to angle adjustments.
For our angles \( A = 45^{\circ} \) and \( B = 60^{\circ} \), we apply the formula by plugging in the known values:

• \( \tan 45^{\circ} = 1 \)
• \( \tan 60^{\circ} = \sqrt{3} \)

Substitute these into the subtraction formula:

• \[ \tan(A - B) = \frac{1 - \sqrt{3}}{1 + 1\sqrt{3}} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \]

Just like in the addition formula, the denominator needs to be rationalized to simplify the expression. This technique ensures our answers are presented in a standardized mathematic form.

Exact Values of Trigonometric Functions

Knowing the exact values of trigonometric functions for common angles is quite essential. These values form the building blocks for more complex trigonometric calculations.

• For example, \( \tan 45^{\circ} = 1 \) is derived from the right triangle properties where the opposite and adjacent are equal.
• Similarly, \( \tan 60^{\circ} = \sqrt{3} \) comes from the special 30-60-90 triangle.

These precise values allow us to handle trigonometric identities more easily, like we've seen with addition and subtraction formulas.
Having these values at your fingertips reduces computational errors and can make solving complex trigonometric equations much simpler.
With exact values under your belt, angles' sum or difference problems no longer need approximations. Thus, simplifying many calculations in mathematics.