Let p = sin35°, q = sin25° and r = sin(-95°). What is (pq + qr + rp) equal to?

Since p + q + r = 0 (from previous), (p+q+r)² = p² + q² + r² + 2(pq+qr+rp) = 0. So pq+qr+rp = -(p²+q²+r²)/2. We need p²+q²+r² = sin²35° + sin²25° + sin²95° = sin²35° + sin²25° + cos²5°. Using sin²35° + cos²35° = 1, etc.: sin²35° = (1-cos70°)/2, sin²25° = (1-cos50°)/2, cos²5° = (1+cos10°)/2. Sum = 3/2 + (-cos70° - cos50° + cos10°)/2. cos70° + cos50° = 2cos60°cos10° = cos10°. So -cos10° + cos10° = 0. Sum = 3/2. Hence pq+qr+rp = -3/4.