What is (\sin 9^{\circ}-\cos 9^{\circ}) equal to?
- A. -\frac{\sqrt{5-\sqrt{5}}}{2} ✓
- B. -\frac{\sqrt{5-\sqrt{3}}}{2}
- C. \frac{\sqrt{5-\sqrt{5}}}{2}
- D. \frac{\sqrt{5-\sqrt{5}}}{4}
Correct Answer: A. -\frac{\sqrt{5-\sqrt{5}}}{2}
Explanation
Let x = \sin 9^{\circ} - \cos 9^{\circ}. Since 9^{\circ} is in the first quadrant and \cos 9^{\circ} \gt \sin 9^{\circ}, x must be negative. Squaring x gives x^2 = \sin^2 9^{\circ} + \cos^2 9^{\circ} - 2\sin 9^{\circ}\cos 9^{\circ} = 1 - \sin 18^{\circ}. We know that \sin 18^{\circ} = \frac{\sqrt{5}-1}{4}. Thus, x^2 = 1 - \frac{\sqrt{5}-1}{4} = \frac{5-\sqrt{5}}{4}. Taking the negative square root gives x = -\frac{\sqrt{5-\sqrt{5}}}{2}.