Problem 46 - Algebra 2 and Trigonometry

Question

Use a counterexample to show that \(\sin A+\sin B=\sin (A+B)\) is false.

Step-by-Step Solution

Answer

\( \sin A + \sin B \neq \sin (A + B) \) when \( A = 30^\circ \) and \( B = 60^\circ \).

Step 1: Understanding the Problem

We need to verify whether the equation \( \sin A + \sin B = \sin (A + B) \) is true for all angles \(A\) and \(B\). A counterexample is sufficient to show that this equation is not always true.

Step 2: Choosing Specific Angles

Let's choose specific angles for \( A \) and \( B \). For simplicity, we will use \( A = 30^\circ \) (or \( \pi/6 \) radians) and \( B = 60^\circ \) (or \( \pi/3 \) radians). These are convenient because their sine values are well known.

Step 3: Calculating \( \sin A + \sin B \)

Calculate \( \sin 30^\circ \) and \( \sin 60^\circ \):\[ \sin 30^\circ = \frac{1}{2} \]\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]Thus, \( \sin A + \sin B = \frac{1}{2} + \frac{\sqrt{3}}{2} \).

Step 4: Calculating \( \sin (A + B) \)

Now, calculate \( \sin (30^\circ + 60^\circ) = \sin 90^\circ \). Knowing that \( \sin 90^\circ = 1 \), we have that \( \sin (A + B) = 1 \).

Step 5: Comparing Results

Now, compare the two results:1. \( \sin A + \sin B = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2} \)2. \( \sin (A + B) = 1 \)Since \( \frac{1 + \sqrt{3}}{2} eq 1 \), the results are not equal.

Key Concepts

Sine Addition Formula

The sine addition formula is one of the essential identities in trigonometry. It's expressed as \( \sin(A + B) = \sin A \cos B + \cos A \sin B \). This identity helps us find the sine of the sum of two angles. It is crucial to verify the sine values by breaking down the angles into their components.

By using the known trigonometric values of usually simpler angles such as 30°, 45°, and 60°, we simplify complex sine calculations. These angles have well-known values, making computation easy. For example:

• \( \sin 30° = \frac{1}{2} \)
• \( \sin 60° = \frac{\sqrt{3}}{2} \)

This identity allows not only for calculations but also ensures formulas align with the fundamental definitions of sine. It guards against misappropriations like believing \( \sin A + \sin B = \sin(A + B) \), which isn’t valid.

Counterexample in Mathematics

Counterexamples play a critical role in mathematics. They are used to demonstrate that a particular statement is false by providing a single example where the statement does not hold.

Consider proving the claim \( \sin A + \sin B = \sin (A + B) \). At first glance, it may seem valid because of apparent symmetry in addition and sine operations. However, when tested against exact angle values, contradictions arise. By choosing angles \( A = 30° \) and \( B = 60° \), we find:

• \( \sin A + \sin B = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2} \)
• \( \sin (A + B) = \sin 90° = 1 \)

Here, the two results are not equal, thus providing a counterexample demonstrating that the initial equation is false. Such counterexamples are powerful tools in verifying the limits of mathematical statements.

Angle Addition in Trigonometry

Angle addition in trigonometry allows us to calculate trigonometric functions for sums of angles efficiently. The key is breaking down complex expressions using identities such as the sine and cosine addition formulas.

When working with angle addition, the rules for operations combined with known angle values simplify the complexity. For instance, for sine, adding two angles A and B, instead of directly using their sine values individually and summing them, the correct identity to use is \( \sin(A + B) = \sin A \cos B + \cos A \sin B \).

This ensures accuracy in calculations involving multiple angles and is especially useful in various applications from physics to engineering. These principles demonstrate not only precision but also the universal application of trigonometry, allowing for consistency in problem-solving across different fields.