Problem 7
Question
In \(3-17,\) find the exact value of \(\cos (A-B)\) for each given pair of values. \(A=270^{\circ}, B=30^{\circ}\)
Step-by-Step Solution
Verified Answer
The exact value is \(-\frac{1}{2}\).
1Step 1: Understand the Problem
We are given angles \(A = 270^{\circ}\) and \(B = 30^{\circ}\) and need to find \(\cos(A - B)\). The formula to use is for the cosine of the difference of two angles: \(\cos(A - B) = \cos A \cos B + \sin A \sin B\).
2Step 2: Use the Identity for Difference of Angles
According to the formula for cosine of the difference of two angles, we need to find \(\cos A\), \(\cos B\), \(\sin A\), and \(\sin B\). Then apply \(\cos(A - B) = \cos A \cos B + \sin A \sin B\).
3Step 3: Find Trigonometric Values
- \(\cos 270^{\circ} = 0\)- \(\sin 270^{\circ} = -1\)- \(\cos 30^{\circ} = \frac{\sqrt{3}}{2}\)- \(\sin 30^{\circ} = \frac{1}{2}\).
4Step 4: Substitute and Calculate
Substitute these values into the identity: \[\cos(270^{\circ} - 30^{\circ}) = \cos 270^{\circ} \cos 30^{\circ} + \sin 270^{\circ} \sin 30^{\circ} \]\[= 0 \times \frac{\sqrt{3}}{2} + (-1) \times \frac{1}{2}\]\[= 0 - \frac{1}{2}\]\[= -\frac{1}{2}\].
5Step 5: Final Result
The value of \(\cos(270^{\circ} - 30^{\circ})\) is \(-\frac{1}{2}\).
Key Concepts
Understanding the Cosine Difference IdentityExact Values of Trigonometric FunctionsAngle Subtraction in Trigonometry
Understanding the Cosine Difference Identity
The cosine of the difference of two angles, often referred to as the cosine difference identity, is a fundamental trigonometric identity. It is given by the formula:
Understanding this identity involves not just memorizing the formula but also recognizing how it can apply in various contexts, such as solving equations or finding exact trigonometric values. Its utility lies in its ability to relate the cosine of a complex angle difference back to basic trigonometric functions.
- \( \cos(A - B) = \cos A \cos B + \sin A \sin B \)
Understanding this identity involves not just memorizing the formula but also recognizing how it can apply in various contexts, such as solving equations or finding exact trigonometric values. Its utility lies in its ability to relate the cosine of a complex angle difference back to basic trigonometric functions.
Exact Values of Trigonometric Functions
In trigonometry, some angles have trigonometric values that we can easily determine without a calculator. These are known as exact values. For example, commonly used angles include \(30^{\circ}\), \(45^{\circ}\), \(60^{\circ}\), \(90^{\circ}\), among others, and their trigonometric functions have set values that are derived from geometric properties or the unit circle.
- For \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \)
- For \( \sin 30^{\circ} = \frac{1}{2} \)
- For \( \cos 270^{\circ} = 0 \)
- For \( \sin 270^{\circ} = -1 \)
Angle Subtraction in Trigonometry
The concept of angle subtraction is essential when dealing with multiple trigonometric identities. When we talk about subtracting angles in trigonometry, we are referring to the concept of taking one angle and removing the measure of another to find their difference.In the context of trigonometric calculations, knowing how to subtract angles allows us to apply identities like the cosine difference identity. For instance, with angles \( A = 270^{\circ} \) and \( B = 30^{\circ} \), subtracting them gives:
- \( A - B = 270^{\circ} - 30^{\circ} = 240^{\circ} \)
Other exercises in this chapter
Problem 7
In \(3-17,\) find the exact value of \(\cos (A+B)\) for each given pair of values. \(A=180^{\circ}, B=30^{\circ}\)
View solution Problem 7
In \(3-14,\) write each expression as a single term using \(\sin \theta, \cos \theta,\) or both. $$ \cot \theta \sec \theta $$
View solution Problem 8
In \(3-8,\) for each value of \(\theta,\) use half-angle formulas to find a. \(\sin \frac{1}{2} \theta\) b. \(\cos \frac{1}{2} \theta\) c. \(\tan \frac{1}{2} \t
View solution Problem 8
In \(3-8,\) for each value of \(\theta,\) use double-angle formulas to find a. \(\sin 2 \theta,\) b. \(\cos 2 \theta,\) c. \(\tan 2 \theta .\) Show all work. $$
View solution