use a double angle formula to solve the equation on the interval [0,2pi)

Question

use a double angle formula or a half angle formula to solve the equation on the interval [0,2pi)

cosθ-sinθ=√2 sin(θ/2)

Answer 1

Hey CJ, to solve this problem I started off by using the half angle formula for sin(θ/2):

sin(θ/2) = ±√((1 - cos(θ)) / 2)

Therefore:

cos(θ) - sin(θ) = √(2)*(±√((1 - cos(θ)) / 2)) Here, all I did is substitute with my half angle formula

(cos(θ) - sin(θ))² = 2 * ((1 - cos(θ)) / 2) Now square both sides, getting rid of both square roots

cos²(θ) + sin²(θ) - 2sin(θ)cos(θ) = 1 - cos(θ) Expand the left side; on the right side the *2 and /2 cancel

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

-2sin(θ)cos(θ) + cos(θ) = 0 Combine everything to one side; the 1's cancel

cos(θ)(-2sin(θ) + 1) = 0 Pull out a cos(θ) because both terms have that in common

cos(θ) = 0 -2sin(θ) +1 = 0 Set both quantities equal to 0

cos(θ) = 0 at π/2 and 3π/2

Solve -2sin(θ) +1 = 0 to get sin(θ) = 1/2 which occurs at π/6 and 5π/6

∴The final answer is θ = π/2, 3π/2, π/6, 5π/6