Chapter 5: Problem 37
Verify the identities. $$ \frac{\cos ^{2} A-\cos ^{2} B}{\cos A+\cos B}=-2\left[\sin \left(\frac{A}{2}\right) \cos \left(\frac{B}{2}\right)+\cos \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right)\right]\left[\sin \left(\frac{A}{2}\right) \cos \left(\frac{B}{2}\right)-\cos \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right)\right] $$
Short Answer
Step by step solution
Apply Pythagorean Identity
Factor Difference of Squares
Rewrite Trigonometric Sum-to-Product Identities
Substitute Identities Back Into Expression
Simplify the Expression
Final Simplification and Verification
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Pythagorean Identity
- \( \cos^2X + \sin^2X = 1 \)
- \( (1 - \sin^2A) - (1 - \sin^2B) = \sin^2B - \sin^2A \)
Difference of Squares
- \( a^2 - b^2 = (a-b)(a+b) \)
In the exercise, we have \( \sin^2B - \sin^2A \), which can be rewritten as the difference of squares. By identifying \( a = \sin B \) and \( b = \sin A \), we can express it as:
- \( (\sin B - \sin A)(\sin B + \sin A) \)
Sum-to-Product Identities
- \( \sin B + \sin A = 2 \sin \left(\frac{B+A}{2}\right) \cos \left(\frac{B-A}{2}\right) \)
- \( \sin B - \sin A = 2 \cos \left(\frac{B+A}{2}\right) \sin \left(\frac{B-A}{2}\right) \)
By applying these transformations, the initially complicated trigonometric expressions become easier to manage and to substitute back into the original equation for further simplification and verification.
Sine and Cosine Functions
- Sine of an angle is defined as the ratio of the length of the opposite side to the hypotenuse.
- Cosine is the ratio of the adjacent side's length to the hypotenuse.
- \( 2 \sin \left(\frac{A}{2}\right) \cos \left(\frac{B}{2}\right) \pm \cos \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \)
Understanding these fundamental trigonometric functions allows us to connect their properties and identities to achieve a simplified result or prove an identity like in the given problem.