What is (pq+qr+rp) equal to?
Consider the following for the three (03) items that follow : Let p=\sin 35^{\circ}, q=\sin 25^{\circ} and r=\sin(-95^{\circ}).
- A. -3/4 ✓
- B. 0
- C. 1/4
- D. 3/4
Correct Answer: A. -3/4
Explanation
Since p+q+r=0, (p+q+r)^2 = 0 \implies p^2+q^2+r^2 + 2(pq+qr+rp) = 0. We evaluate p^2+q^2+r^2 = \sin^2 35^{\circ} + \sin^2 25^{\circ} + \cos^2 5^{\circ} = \frac{1-\cos 70^{\circ}}{2} + \frac{1-\cos 50^{\circ}}{2} + \cos^2 5^{\circ} = 1 - \frac{1}{2}(\cos 70^{\circ} + \cos 50^{\circ}) + \cos^2 5^{\circ} = 1 - \cos 60^{\circ}\cos 10^{\circ} + \cos^2 5^{\circ} = 1 - \frac{1}{2}(2\cos^2 5^{\circ} - 1) + \cos^2 5^{\circ} = \frac{3}{2}. Hence, 2(pq+qr+rp) = -3/2 \implies pq+qr+rp = -3/4.