What is a value of \cos^{3}x+\cos^{3}y?
Consider the following for the next two (02) items that follow : Let \sin x+\sin y=\sqrt{3}(\cos y-\cos x); x+y=\frac{\pi}{2}, 0 \lt x, y \lt \frac{\pi}{2}.
- A. \frac{3\sqrt{3}}{8}
- B. \frac{3\sqrt{6}}{8} ✓
- C. \frac{3\sqrt{6}}{4}
- D. 1
Correct Answer: B. \frac{3\sqrt{6}}{8}
Explanation
Using the specific angles x = 75^\circ and y = 15^\circ determined in the previous problem, we find \cos^3 75^\circ + \cos^3 15^\circ = (\cos 75^\circ + \cos 15^\circ)(\cos^2 75^\circ - \cos 75^\circ \cos 15^\circ + \cos^2 15^\circ). Using sum-to-product, \cos 75^\circ + \cos 15^\circ = \frac{\sqrt{6}}{2}. The second factor evaluates to 1 - \frac{1}{4} = \frac{3}{4}. The product is \left(\frac{\sqrt{6}}{2}\right)\left(\frac{3}{4}\right) = \frac{3\sqrt{6}}{8}.