sin2 (120° - A) + sin2 A + sin2 (120+A) is equal to ? |
\(\frac{3}{2}\) \(\frac{5}{16}\) \(\frac{1}{3}\) \(\frac{7}{16}\) |
\(\frac{3}{2}\) |
Let A = 60° sin2 (120°-60°) + sin2 (60°) + sin2 (180°) = sin2 60° + sin2 60° + sin2 180° = \(\frac{3}{4}\) × 2 + 0 = \(\frac{3}{2}\) |