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 with π/2 - θ, and the fraction becomes a single e^(iα).

1. Change the angle. 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.
2. Half-angle form. Use 1 + cos α = 2 cos² α/2 and sin α = 2 sin α/2 cos α/2. The numerator becomes 2 cos α/2 e^(iα/2). The denominator is its conjugate, 2 cos α/2 e^(-iα/2).

   (e^(iα/2)/e^(-iα/2))³ = (e^(iα))³ = e^(i 15π/18) = e^(i 5π/6)

3. Evaluate. By De Moivre, e^(i5π/6) = cos 150° + i sin 150° = -√3/2 + i/2.
4. So. The value is -1/2(√3 - i), which is option (b).

The idea to keep. For (1 + cos α + i sin α)/(1 + cos α - i sin α), the answer is always e^(iα). If the sine and cosine sit in the wrong places, switch to π/2 - θ first.

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