Problem 5 - 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. $$ A=60^{\circ}, B=60^{\circ} $$

Step-by-Step Solution

Answer

\(\tan(120^{\circ}) = -\sqrt{3}\) and \(\tan(0^{\circ}) = 0\).

Step 1: Determine individual tangent values

First, we need to calculate the tangent of each angle. Since both angles are \(60^{\circ}\), we know that \(\tan(60^{\circ}) = \sqrt{3}\).

Step 2: Add and subtract angles

Calculate \(A+B\) and \(A-B\). Here, \(A+B = 60^{\circ} + 60^{\circ} = 120^{\circ}\) and \(A-B = 60^{\circ} - 60^{\circ} = 0^{\circ}\).

Step 3: Apply tangent angle sum formula

For \(\tan(A+B)\), use the identity \(\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\). Substitute \(\tan(60^{\circ}) = \sqrt{3}\), resulting in:\[\tan(120^{\circ}) = \frac{\sqrt{3} + \sqrt{3}}{1 - \sqrt{3} \times \sqrt{3}} = \frac{2\sqrt{3}}{1 - 3} = -\sqrt{3}\]

Step 4: Apply tangent angle difference formula

For \(\tan(A-B)\), use the identity \(\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\). With both angles equal, substitute again:\[\tan(0^{\circ}) = \frac{\sqrt{3} - \sqrt{3}}{1 + \sqrt{3}\times\sqrt{3}} = \frac{0}{4} = 0\]

Key Concepts

Tangent of Angles

The tangent of an angle in trigonometry is defined using the sides of a right triangle. Specifically, for an angle \(\theta\), the tangent is the ratio of the length of the opposite side to the adjacent side. This can be expressed as \( \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \).

In the unit circle, where the hypotenuse is 1, tangent can also be understood as the ratio of sine to cosine: \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \).

Different angles have specific tangent values. For example:

• \( \tan(0^\circ) = 0 \)
• \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \)
• \( \tan(45^\circ) = 1 \)
• \( \tan(60^\circ) = \sqrt{3} \)

The tangent function is periodic, repeating every \(180^\circ\) or \(\pi\) radians, and is undefined when the cosine of the angle equals zero.

Angle Addition and Subtraction

When dealing with trigonometric functions like tangent, combining angles can be calculated using specific identities. These identities are essential for simplifying expressions and solving equations.

*Angle Addition Formula*
The formula for the tangent of the sum of two angles \(A\) and \(B\) is:\[\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\]

*Angle Subtraction Formula*
Similarly, the tangent of the difference between two angles is given by:\[\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\]

These formulas help calculate the tangent of new angles formed by adding or subtracting angles with known tangent values.

In our example with \(A = 60^\circ\) and \(B = 60^\circ\):

• \(A+B = 120^\circ\)
• \(A-B = 0^\circ\)

We used these formulas to find the exact values for \(\tan(120^\circ)\) and \(\tan(0^\circ)\).

Trigonometric Formulas

Trigonometric formulas are the backbone of solving trigonometry problems involving angles and their relationships. They provide critical identities and relationships for computing the trigonometric functions of various angles.

Some important formulas include:

• **Pythagorean Identity**: \( \sin^2(\theta) + \cos^2(\theta) = 1 \)
• **Quotient Identities**: \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \) and \( \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} \)
• **Even-Odd Identities**: \( \tan(-\theta) = -\tan(\theta) \)
• **Double Angle Formulas**: Especially for tangent, \( \tan(2\theta) = \frac{2\tan(\theta)}{1-\tan^2(\theta)} \)

These formulas simplify the process of finding the tangent and other trigonometric functions for different angles.

Understanding these relationships is crucial for tackling more complex trigonometry problems and real-life applications, such as navigating by angles and calculating the slope in engineering and physics.