Problem 12
Question
In \(3-17,\) find the exact value of \(\cos (A-B)\) for each given pair of values. \(A=45^{\circ}, B=270^{\circ}\)
Step-by-Step Solution
Verified Answer
The exact value of \( \cos(45^{\circ} - 270^{\circ}) \) is \(-\frac{\sqrt{2}}{2}\).
1Step 1: Recall the Cosine Difference Identity
To find \( \cos(A-B) \), use the identity: \( \cos(A-B) = \cos A \cdot \cos B + \sin A \cdot \sin B \).
2Step 2: Determine Cosine and Sine of Given Angles
For \( A = 45^{\circ} \), we have \( \cos A = \frac{\sqrt{2}}{2} \) and \( \sin A = \frac{\sqrt{2}}{2} \).For \( B = 270^{\circ} \), we have \( \cos B = 0 \) and \( \sin B = -1 \).
3Step 3: Substitute Values into the Identity
Substitute the values found in Step 2 into \( \cos(A-B) = \cos A \cdot \cos B + \sin A \cdot \sin B \):\[ \cos(45^{\circ} - 270^{\circ}) = \left(\frac{\sqrt{2}}{2}\right) \cdot 0 + \left(\frac{\sqrt{2}}{2}\right) \cdot (-1) \]
4Step 4: Simplify the Expression
Simplify the expression:\[ \cos(45^{\circ} - 270^{\circ}) = 0 - \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2} \]
5Step 5: Final Step: Provide the Exact Value
Thus, the exact value of \( \cos(A-B) \) for \( A = 45^{\circ} \) and \( B = 270^{\circ} \) is \(-\frac{\sqrt{2}}{2}\).
Key Concepts
Trigonometric IdentitiesCosine and Sine ValuesAngle Subtraction
Trigonometric Identities
Trigonometric identities are mathematical tools that relate the angles and lengths of a triangle. These identities are essential in solving a wide range of problems in trigonometry. One such useful identity is the Cosine Difference Identity. This identity helps find the cosine of the difference between two angles. It states:
- \[ \cos(A-B) = \cos A \cdot \cos B + \sin A \cdot \sin B\]
Cosine and Sine Values
Accurate determination of cosine and sine values for specific angles is vital in trigonometry. These values are fundamental in using trigonometric identities. Often, angles such as \(45^{\circ}\), \(90^{\circ}\), and \(270^{\circ}\) are encountered, where knowing their trigonometric function values can drastically simplify calculations.For instance:
- For \(A = 45^{\circ}\), the values are:
\( \cos A = \frac{\sqrt{2}}{2} \)
\( \sin A = \frac{\sqrt{2}}{2} \) - For \(B = 270^{\circ}\), the values are:
\( \cos B = 0 \)
\( \sin B = -1 \)
Angle Subtraction
Angle subtraction is a critical calculation in trigonometry. When we are asked to find \(\cos(A-B)\), it means we want to determine the cosine of the difference between two angles, \(A\) and \(B\). Here, angle subtraction becomes crucial in using the Cosine Difference Identity, as it helps to find the combined effect of these angles.By applying the identity:
\[ \cos(45^{\circ} - 270^{\circ}) = \left(\frac{\sqrt{2}}{2}\right) \cdot 0 + \left(\frac{\sqrt{2}}{2}\right) \cdot (-1) \]
Which results in:
\[ 0 - \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2} \]By understanding angle subtraction and applying trigonometric identities accurately, the exact value of complex expressions can be found efficiently.
- \( \cos(45^{\circ} - 270^{\circ}) \)
- \( \cos A = \frac{\sqrt{2}}{2}, \sin A = \frac{\sqrt{2}}{2} \)
- \( \cos B = 0, \sin B = -1 \)
\[ \cos(45^{\circ} - 270^{\circ}) = \left(\frac{\sqrt{2}}{2}\right) \cdot 0 + \left(\frac{\sqrt{2}}{2}\right) \cdot (-1) \]
Which results in:
\[ 0 - \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2} \]By understanding angle subtraction and applying trigonometric identities accurately, the exact value of complex expressions can be found efficiently.
Other exercises in this chapter
Problem 12
In \(3-17,\) find the exact value of \(\cos (A+B)\) for each given pair of values. \(A=60^{\circ}, B=60^{\circ}\)
View solution Problem 12
In \(3-14,\) write each expression as a single term using \(\sin \theta, \cos \theta,\) or both. $$ \frac{\tan \theta}{\cot \theta}+\tan \theta \cot \theta $$
View solution Problem 13
In \(3-17,\) find the exact value of \(\tan (A+B)\) and of \(\tan (A-B)\) for each given pair of values. $$ A=240^{\circ}, B=120^{\circ} $$
View solution Problem 13
In \(3-26,\) prove that each equation is an identity. $$ \frac{\cot \theta}{\csc \theta}=\cos \theta $$
View solution