Problem 6 - Algebra 2 and Trigonometry

Question

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

Step-by-Step Solution

Answer

\(\cos(210^{\circ}) = -\frac{\sqrt{3}}{2}\)

Step 1: Recall the cosine of difference formula

The formula for the cosine of the difference between two angles, \(A\) and \(B\), is given by: \[ \cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B \] We'll use this formula to compute \(\cos(270^{\circ} - 60^{\circ})\).

Step 2: Calculate \(A - B\)

Subtract the angle \(B = 60^{\circ}\) from \(A = 270^{\circ}\):\[ A - B = 270^{\circ} - 60^{\circ} = 210^{\circ} \]Now we need to find \(\cos(210^{\circ})\).

Step 3: Find exact values of trigonometric functions

We will use the known trigonometric values:- \(\cos(210^{\circ}) = -\cos(30^{\circ}) = -\frac{\sqrt{3}}{2}\)- \(\sin(210^{\circ}) = -\sin(30^{\circ}) = -\frac{1}{2}\)These come from considering that 210° is in the third quadrant, where cosine and sine are both negative.

Step 4: Plug values into the formula

Using the cosine of a difference formula:\[ \cos(210^{\circ}) = \cos(270^{\circ} - 60^{\circ}) = \cos(270^{\circ}) \cdot \cos(60^{\circ}) + \sin(270^{\circ}) \cdot \sin(60^{\circ}) \]

Step 5: Evaluate trigonometric functions at special angles

We need the trigonometric values for the angles involved:- \(\cos(270^{\circ}) = 0\)- \(\sin(270^{\circ}) = -1\)- \(\cos(60^{\circ}) = \frac{1}{2}\)- \(\sin(60^{\circ}) = \frac{\sqrt{3}}{2}\)

Step 6: Calculate the expression

Substitute the values calculated into the formula:\[ \cos(270^{\circ}) \cdot \cos(60^{\circ}) + \sin(270^{\circ}) \cdot \sin(60^{\circ}) = 0 \cdot \frac{1}{2} + (-1) \cdot \frac{\sqrt{3}}{2} \]This simplifies to:\[ 0 - \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2} \]

Key Concepts

Cosine of a Difference

When working with angles, often you'll need to calculate the cosine of the difference between two angles. This is where the cosine of difference identity becomes incredibly useful. The formula is:\[ \cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B \]This formula helps you find the cosine of the difference between two angles by using their individual sine and cosine values. For example, if you have angles \(A = 270^{\circ}\) and \(B = 60^{\circ}\), you can find \(\cos(210^{\circ})\) by evaluating:

• \(\cos(270^{\circ})\cdot\cos(60^{\circ})\)
• \(\sin(270^{\circ})\cdot\sin(60^{\circ})\)

Each term sums up to the exact value of the cosine of the difference between the two angles.

Sine and Cosine in Quadrants

Trigonometric functions change their signs depending on the quadrant they fall into. Here's what you should remember:

• In the first quadrant, both \(\sin\) and \(\cos\) are positive.
• In the second quadrant, \(\sin\) is positive while \(\cos\) is negative.
• In the third quadrant, both \(\sin\) and \(\cos\) are negative.
• In the fourth quadrant, \(\sin\) is negative while \(\cos\) is positive.

For example, \(210^{\circ}\) is located in the third quadrant where both sine and cosine values are negative. Understanding the quadrant helps you quickly determine the sign of trigonometric values, like knowing that \(\cos(210^{\circ}) = -\frac{\sqrt{3}}{2}\).

Exact Trigonometric Values

Certain angles have known exact trigonometric values. These include angles like \(30^{\circ}, 45^{\circ}, 60^{\circ}, 90^{\circ}\), and similar angles in other quadrants obtained by adding or subtracting multiples of \(180^{\circ}\). Knowing these exact values allows for quick calculations without needing a calculator. Here are some of the key exact values to remember:

• \(\cos(60^{\circ}) = \frac{1}{2}\)
• \(\sin(60^{\circ}) = \frac{\sqrt{3}}{2}\)
• \(\cos(30^{\circ}) = \frac{\sqrt{3}}{2}\)
• \(\sin(30^{\circ}) = \frac{1}{2}\)

By understanding these fundamental values, you can solve problems involving angle differences, like finding \(\cos(210^{\circ})\), much more easily. Recognizing and using these values efficiently is key to mastering trigonometry basics.