Pre calculus question

Question

Use a Double- or Half-Angle Formula to solve the equation in the interval [0, 2pi). (Enter your answers as a comma-separated list.)

cos(theta) - sin(theta) = square root 2 sin(theta/2)

Answer 1

cosθ - sinθ = √2sin(θ/2);

If we square both side: 1 - 2cosθsinθ = 1 - cosθ; cosθ - 2cosθsinθ = 0; cosθ(1 - 2sinθ) = 0

cosθ = 0; θ = π/2 or θ = 3π/2; 1 - 2sinθ = 0; sinθ = 1/2; θ = π/6 or θ = 5π/6

Check: x = π/2 cosπ/2 - sinπ/2 = √2sinπ/4; 0 - 1 = 1 contradiction

x = 3π/2; cos3π/2 - sin3π/2 = √2sin3π/4; 0 + 1 = 1 OK

x = π/6; cosπ/6 - sinπ/6 = √2sinπ/12; (√3 - 1)/2 = √2·√(1 - √3/2)/2 = √[(2 - √3)/2} OK

x = 5π/6; cos5π/6 - sin5π/6 = √2sin5π/12; - √3/2 - 1/2 = √2sin5π/12 contradiction (negative = positive)

Answer: x = 3π/2, x = π/6

Answer 2

cos(θ)-sin(θ)=√2sin(θ/2)=√(2sin²(θ/2)) = √(2((1-cos(θ))/2)

Square both sides

(cos(θ)-sin(θ))² = 1-cos(θ)

sin²(θ)+cos²(θ)-2sin(θ)cos(θ) = 1-cos(θ)

1-2sin(θ)cos(θ) = 1-cos(θ)

2sin(θ)cos(θ) = cos(θ)

2sin(θ)cos(θ) - cos(θ)=0

cos(θ)(2sin(θ)-1) = 0

cos(θ) = 0 --> θ=π/2,3π/2

sin(θ)=1/2 --> θ=π/6, 11π/6

θ = π/2, 3π/2, π/6, 11π/6

Answer 3

cos θ - sin θ = (√2)(±√(1 - cos θ) / √2)

Simplify right side.

Square both sides

Cobine

Use Zero Product Rule