Question

cos\(^6\) A - sin\(^6\) A is equal to

• A. \(\cos 2A \left( 1 - \frac{1}{4} \sin^2 2A \right)\)
• B. \(\cos 2A \left( 1 - \frac{3}{4} \sin^2 2A \right)\)
• C. \(\cos 2A \left( 1 - \frac{1}{2} \sin^2 2A \right)\)
• D. \(\cos 2A \left( 1 + \frac{1}{4} \sin^2 2A \right)\)

Hint:

For simplifying powers of trigonometric functions, look for opportunities to apply standard identities such as the double angle formulas and Pythagorean identities.

Answer 1

The Correct Option is A

We are given the expression \( \cos^6 A - \sin^6 A \), which is a difference of cubes. We can factor it using the identity: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \] Thus, we can factor \( \cos^6 A - \sin^6 A \) as: \[ \cos^6 A - \sin^6 A = (\cos^2 A - \sin^2 A) \left( \cos^4 A + \cos^2 A \sin^2 A + \sin^4 A \right) \] We know that \( \cos^2 A - \sin^2 A = \cos 2A \), so the expression becomes: \[ \cos^6 A - \sin^6 A = \cos 2A \left( \cos^4 A + \cos^2 A \sin^2 A + \sin^4 A \right) \] Now, we simplify the second factor. Using the identity \( \cos^2 A + \sin^2 A = 1 \), we rewrite \( \cos^4 A + \sin^4 A \) as: \[ \cos^4 A + \sin^4 A = (\cos^2 A + \sin^2 A)^2 - 2\cos^2 A \sin^2 A = 1 - 2\cos^2 A \sin^2 A \] Thus, the second factor becomes: \[ \cos^4 A + \cos^2 A \sin^2 A + \sin^4 A = 1 - 2\cos^2 A \sin^2 A + \cos^2 A \sin^2 A = 1 - \cos^2 A \sin^2 A \] Finally, we recognize that \( \cos^2 A \sin^2 A = \frac{1}{4} \sin^2 2A \), so the expression simplifies to: \[ \cos^6 A - \sin^6 A = \cos 2A \left( 1 - \frac{1}{4} \sin^2 2A \right) \] Thus, the correct answer is \( \cos 2A \left( 1 - \frac{1}{4} \sin^2 2A \right) \).