Problem 11 - Algebra 2 and Trigonometry

Question

\(\ln 3-17,\) find the exact value of \(\sin (A-B)\) and of \(\sin (A+B)\) for each given pair of values. \(A=60^{\circ}, B=270^{\circ}\)

Step-by-Step Solution

Answer

\(\sin(A-B) = \frac{1}{2}\); \(\sin(A+B) = -\frac{1}{2}\)

Step 1: Understand the Problem

We need to find the exact values of \(\sin(A-B)\) and \(\sin(A+B)\) given the angles \(A=60^{\circ}\) and \(B=270^{\circ}\). These involve applying the sine subtraction and addition formulas.

Step 2: Recall the Sine Subtraction Formula

The sine subtraction formula is \(\sin(A-B) = \sin A \cos B - \cos A \sin B\). We will use this formula to find \(\sin(60^{\circ} - 270^{\circ})\).

Step 3: Recall the Sine Addition Formula

The sine addition formula is \(\sin(A+B) = \sin A \cos B + \cos A \sin B\). We will use this formula to find \(\sin(60^{\circ} + 270^{\circ})\).

Step 4: Calculate \(\sin 60^{\circ}\) and \(\cos 60^{\circ}\)

We know that \(\sin 60^{\circ} = \frac{\sqrt{3}}{2}\) and \(\cos 60^{\circ} = \frac{1}{2}\). These will be used in the formulas.

Step 5: Calculate \(\sin 270^{\circ}\) and \(\cos 270^{\circ}\)

We know that \(\sin 270^{\circ} = -1\) and \(\cos 270^{\circ} = 0\). These values are needed to apply the formulas.

Step 6: Apply Sine Subtraction Formula

Using the formula: \(\sin(60^{\circ} - 270^{\circ}) = \sin 60^{\circ} \cos 270^{\circ} - \cos 60^{\circ} \sin 270^{\circ} = \frac{\sqrt{3}}{2} \times 0 - \frac{1}{2} \times (-1) = \frac{1}{2}\).

Step 7: Apply Sine Addition Formula

Using the formula: \(\sin(60^{\circ} + 270^{\circ}) = \sin 60^{\circ} \cos 270^{\circ} + \cos 60^{\circ} \sin 270^{\circ} = \frac{\sqrt{3}}{2} \times 0 + \frac{1}{2} \times (-1) = -\frac{1}{2}\).

Key Concepts

Sine Subtraction Formula

The sine subtraction formula is an essential tool in trigonometry. It enables the calculation of the sine of the difference between two angles. This formula expresses

• the sine of \( A-B \) as: \[ \sin(A - B) = \sin A \cos B - \cos A \sin B \]

The formula is derived based on the unit circle and properties of sine and cosine. It holds true for any angles, measured in degrees or radians.

In our exercise, using the values for \(A = 60^{\circ} \) and \( B = 270^{\circ}\), we substitute directly into the formula:

• Find \( \sin 60^{\circ}\) and \( \cos 60^{\circ}\) which are \( \frac{\sqrt{3}}{2} \) and \( \frac{1}{2} \) respectively.
• Find \( \sin 270^{\circ}\) and \( \cos 270^{\circ}\) which are -1 and 0 respectively.

Using these values, the formula becomes: \[ \sin (60^{\circ} - 270^{\circ}) = \frac{\sqrt{3}}{2} \times 0 - \frac{1}{2} \times (-1) = \frac{1}{2} \] This illustrates how the formula simplifies the calculation of sine for angle differences.

Sine Addition Formula

The sine addition formula is the counterpart to the subtraction formula, used to find the sine of two added angles. It is given by:

• \[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]

This formula is particularly useful in simplifying the trigonometric expression for \( \sin (A + B) \).

In the case of the exercise with angles \( A = 60^{\circ} \) and \( B = 270^{\circ} \), we follow similar steps:

• Substitute \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \) and \( \cos 60^{\circ} = \frac{1}{2} \).
• For \( 270^{\circ} \), \( \cos 270^{\circ} = 0 \) and \( \sin 270^{\circ} = -1 \).

Plugging these into our formula, we calculate:\[ \sin(60^{\circ} + 270^{\circ}) = \frac{\sqrt{3}}{2} \times 0 + \frac{1}{2} \times (-1) = -\frac{1}{2} \] This highlights how the sine addition formula can be used for precise trigonometric calculations.

Angle Subtraction and Addition

Understanding angle subtraction and addition is fundamental in trigonometry. These concepts explain how to manipulate angles and apply trigonometric formulas.The subtraction and addition of angles are crucial for:

• Reducing complex trigonometric expressions.
• Solving trigonometric equations.
• Analyzing rotating systems and wave patterns.

In our context, using \( A = 60^{\circ} \) and \( B = 270^{\circ} \), subtraction was employed to find \( \sin(A - B)\), which resulted in \( \frac{1}{2} \).

Addition was used to obtain \( \sin(A + B)\), which gave \(-\frac{1}{2}\). Understanding and applying these processes ensures that students gain deep insight into how simple calculations are connected to broader trigonometric concepts.