The value of
( (1 + sin 2π/9 + i cos 2π/9)/(1 + sin 2π/9 - i cos 2π/9) )³ is :

Solution

Maths sir · 4 steps

Turn sin into cos of the complementary angle, and the fraction becomes a single e^(iα).

1. Switch to complement. Let α = π/2 - 2π/9 = 5π/18. Then sin 2π/9 = cos α and cos 2π/9 = sin α, so the numerator is 1 + cos α + i sin α and the denominator is 1 + cos α - i sin α.
2. Half-angle form. Use 1 + cos α = 2 cos² α/2 and sin α = 2 sin α/2 cos α/2. The common factor 2 cos α/2 cancels from top and bottom.

   (1+cos α + i sin α)/(1+cos α - i sin α) = (cos α/2 + i sin α/2)/(cos α/2 - i sin α/2) = e^(iα/2)/e^(-iα/2) = e^(iα)

3. Cube it. By De Moivre, the cube is e^(3iα) = e^(i· 15π/18) = e^(i· 5π/6). That is cos 150° + i sin 150° = -√3/2 + i/2.
4. So. -√3/2 + i/2 = -1/2(√3 - i), which is option (b).

The idea to keep. When you see 1 + cos α ± i sin α, write it as 2 cos α/2 e^(± iα/2). If sin and cos appear swapped, first move to the complementary angle.

Answer from the NTA final answer key. Solution by PrepWiser.