Question

If \(\cos \alpha + \cos \beta + \cos \gamma = \sin \alpha + \sin \beta + \sin \gamma = 0,\) then evaluate \((\cos^3 \alpha + \cos^3 \beta + \cos^3 \gamma)^2 + (\sin^3 \alpha + \sin^3 \beta + \sin^3 \gamma)^2 =\)

• A. \( 1 \)
• B. \( \frac{3}{4} \)
• C. \( \frac{9}{16} \)
• D. \( \frac{9}{8} \)

Hint:

For problems involving trigonometric sums and cubes, use algebraic identities and sum-to-product transformations to simplify the expressions systematically.

Answer 1

The Correct Option is C

We are given the conditions: \[ \cos \alpha + \cos \beta + \cos \gamma = 0, \quad \sin \alpha + \sin \beta + \sin \gamma = 0. \] Step 1: Use of Identity Using the identity for cubes: \[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx). \] Since \( \cos \alpha + \cos \beta + \cos \gamma = 0 \) and \( \sin \alpha + \sin \beta + \sin \gamma = 0 \), we substitute: \[ \cos^3 \alpha + \cos^3 \beta + \cos^3 \gamma = 3 \cos \alpha \cos \beta \cos \gamma. \] \[ \sin^3 \alpha + \sin^3 \beta + \sin^3 \gamma = 3 \sin \alpha \sin \beta \sin \gamma. \] Step 2: Square and Sum \[ (\cos^3 \alpha + \cos^3 \beta + \cos^3 \gamma)^2 + (\sin^3 \alpha + \sin^3 \beta + \sin^3 \gamma)^2 \] \[ = 9 (\cos^2 \alpha \cos^2 \beta \cos^2 \gamma + \sin^2 \alpha \sin^2 \beta \sin^2 \gamma). \] Using trigonometric identities: \[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = \frac{3}{4}, \quad \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = \frac{3}{4}. \] \[ \cos^2 \alpha \cos^2 \beta \cos^2 \gamma + \sin^2 \alpha \sin^2 \beta \sin^2 \gamma = \frac{1}{16}. \] Step 3: Compute Final Value \[ 9 \times \frac{1}{16} = \frac{9}{16}. \] Thus, the correct answer is: \[ \boxed{\frac{9}{16}} \]