Find all the values of θ between 0 and 2π for which

Question

cos(θ/2) = √(1 + sinθ) - √(1 - sinθ) holds true?

Answer 1

The equation is cos(θ/2) = √[(1 + sin(θ)] - √[(1 - sin(θ)] and the interval is [0, 2π] and solve for θ.

Square on both sides.

cos²(θ/2) = {√[(1 + sin(θ)] - √[(1 - sin(θ)]}²

cos²(θ/2) = (1 + sin(θ) + (1 - sin(θ) - 2[√(1 - sin(θ)²(1 - sin(θ)²].

cos²(θ/2) = (1 + sin(θ) + (1 - sin(θ) - 2[√(1 - sin(θ)(1 + sin(θ)]

cos²(θ/2) = 2 - 2[√(1 - sin²(θ)]

cos²(θ/2) = 2 - 2cos(θ)

cos²(θ/2) = 2 - 2cos[2(θ/2)]

cos²(θ/2) = 2 - 2[2cos²(θ/2) - 1]

cos²(θ/2) = 2 - 4cos²(θ/2) + 2

5cos²(θ/2) = 4

cos²(θ/2) = 4/5

cos(θ/2) = ± √(4/5)

cos(θ/2) = ± cos(26.57).

Let θ/2 = t ⇒ cos(t) = cos(26.57^o).

General solution: θ = 2nπ ± α, where n is an integer.

If α = 26.57^o ⇒ t = 360^on ± 26.57^o ⇒ θ/2 = 360^on ± 26.57^o ⇒ θ = 720^on ± 53.13^o.

If n = 0 ⇒ θ = 720^on ± 53.13^o = 53.13^o.

The solution in the interval [0, 2π] is 53.13^o.