Problem 19 - Algebra 2 and Trigonometry

Question

Find \(\tan (A+B)\) if \(\tan A=3\) and \(\tan B=-\frac{1}{2}\)

Step-by-Step Solution

Answer

\( \tan(A+B) = 1 \)

Step 1: Understand the Problem

We are given that \( \tan A = 3 \) and \( \tan B = -\frac{1}{2} \). We need to find \( \tan (A+B) \). This involves using the formula for the tangent of a sum of angles.

Step 2: Recall the Formula

The formula to find \( \tan (A + B) \) is \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B} \). We'll apply this formula using the given values.

Step 3: Substitute Values into the Formula

Substitute \( \tan A = 3 \) and \( \tan B = -\frac{1}{2} \) into the formula: \[\tan(A+B) = \frac{3 + \left(-\frac{1}{2}\right)}{1 - 3 \times \left(-\frac{1}{2}\right)}\]

Step 4: Simplify the Numerator

Calculate the numerator of the fraction:\[ 3 + \left(-\frac{1}{2}\right) = 3 - \frac{1}{2} = \frac{6}{2} - \frac{1}{2} = \frac{5}{2} \]

Step 5: Simplify the Denominator

Calculate the denominator of the fraction:\[ 1 - 3 \times \left(-\frac{1}{2}\right) = 1 + \frac{3}{2} = \frac{2}{2} + \frac{3}{2} = \frac{5}{2} \]

Step 6: Calculate \( \tan(A+B) \)

Using the simplified numerator and denominator, calculate \( \tan(A+B) \):\[\tan(A+B) = \frac{\frac{5}{2}}{\frac{5}{2}} = 1\]

Step 7: Finalize the Result

From the calculation, \( \tan(A+B) = 1 \). Thus, the tangent of the sum of angles \( A \) and \( B \) is 1.

Key Concepts

Tangent of Sum of Angles

Trigonometry offers a variety of identities to uncover unknown angles and their relationships. The tangent of a sum of angles is particularly useful. This specific identity helps calculate the tangent of the sum of two angles when you know the tangents of the individual angles. The formula is: \[ \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B} \] It allows us to dissect complex calculations into simpler ones by breaking them down into known values. For any angles \(A\) and \(B\), all you need are their respective tangent values. Then plug those into the formula to find the desired tangent. This adds a layer of structured calculation, making it handy in solving a plethora of geometric problems. Understanding and mastering this formula can significantly ease the process of working through angle-related exercises.

Basic Trigonometric Functions

Trigonometric functions—sine, cosine, and tangent—are fundamental building blocks in trigonometry. Each function represents specific ratios of sides in a right-angled triangle. Here's an overview:

• Sine (\( \sin \)): The ratio of the length of the opposite side to the hypotenuse.
• Cosine (\( \cos \)): The ratio of the length of the adjacent side to the hypotenuse.
• Tangent (\( \tan \)): The ratio of the length of the opposite side to the adjacent side. \( \tan x = \frac{\sin x}{\cos x} \).

Familiarity with these ratios is crucial because they form the principles underpinning many trigonometric problems. For instance, in our problem, knowing \(\tan A\) and \(\tan B\) is essential to implementing the tangent sum formula. By breaking down complex angles into these functions, solving trigonometric equations becomes a manageable endeavor.

Algebraic Manipulation

Algebraic manipulation plays a vital role in trigonometry, helping to simplify expressions and solve equations. In the context of the exercise, we use algebra to break down and solve the identity for \(\tan(A + B)\). It's about applying algebraic rules, such as:

• Simplifying fractions: Combine, subtract, or divide terms correctly to reach a simpler form.
• Distributive properties: Handle expressions involving multiplication over addition or subtraction effectively.
• Substitution: Place known values into formulas to compute unknowns successfully.

In our example, substituting \(\tan A = 3\) and \(\tan B = -\frac{1}{2}\) into the formula necessitated careful arithmetic on both the numerator and denominator. Simplifying these components step by step lead to the final, simplified term \(\tan(A+B) = 1\). This highlights the seamless blend of trigonometry and algebra required to tackle such problems proficiently.