Chapter 25: Problem 59
Let \(\alpha, \beta\) be such that \(\pi<\alpha-\beta<3 \pi\). If \(\sin \alpha+\sin \beta=-\frac{21}{65}\) and \(\cos \alpha+\cos \beta=-\frac{27}{65}\), then the value of \(\cos \frac{\alpha-\beta}{2}\) is (A) \(-\frac{3}{\sqrt{130}}\) (B) \(\frac{3}{\sqrt{130}}\) (C) \(\frac{6}{65}\) (D) \(-\frac{6}{65}\)
Short Answer
Step by step solution
Use Sum-to-Product Identities
Form Equations Using Given Values
Isolate \(\cos \left( \frac{\alpha-\beta}{2} \right)\)
Use Pythagorean Identity
Solve for \(x\)
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sum-to-Product Identities
- \( \sin \alpha + \sin \beta = 2 \sin \left( \frac{\alpha+\beta}{2} \right) \cos \left( \frac{\alpha-\beta}{2} \right) \)
- \( \cos \alpha + \cos \beta = 2 \cos \left( \frac{\alpha+\beta}{2} \right) \cos \left( \frac{\alpha-\beta}{2} \right) \)
Trigonometric Equations
In this exercise, once we substitute the values into the sum-to-product identities, we form two separate equations:
- \( 2 \sin \left( \frac{\alpha+\beta}{2} \right) \cos \left( \frac{\alpha-\beta}{2} \right) = -\frac{21}{65} \)
- \( 2 \cos \left( \frac{\alpha+\beta}{2} \right) \cos \left( \frac{\alpha-\beta}{2} \right) = -\frac{27}{65} \)
Pythagorean Identity
- \( \sin^2 \theta + \cos^2 \theta = 1 \)
In the current problem, the Pythagorean Identity is utilized after isolating \( x = \cos \left( \frac{\alpha-\beta}{2} \right) \). We reached the equation:
- \( x^2 \left( \sin^2 \left( \frac{\alpha+\beta}{2} \right) + \cos^2 \left( \frac{\alpha+\beta}{2} \right) \right) = \left( \frac{21}{130} \right)^2 + \left( \frac{27}{130} \right)^2 \)