Find the points on the given curve where the tangent line is horizontal or vertical. (Assume 0 ≤ θ ≤ 2π. Enter your answers as a comma-separated list of ordered pairs.) r = 1 − sin(θ) horizontal tangent

The tangent to the curve at a point P (x, y) has slope dy/dx at that point. By the chain rule,

dy/dx = (dy/dθ) / (dx/dθ)

We're in polar coordinates, so

y (θ) = r (θ) sin(θ)   ==>   dy/dθ = dr/dθ sin(θ) + r (θ) cos(θ)

x (θ) = r (θ) cos(θ)   ==>   dx/dθ = dr/dθ cos(θ) - r (θ) sin(θ)

We're given r (θ) = 1 - sin(θ), so that

dr/dθ = -cos(θ)

Then the slope of the tangent to the curve at P is

dy/dx = (dr/dθ sin(θ) + r (θ) cos(θ)) / (dr/dθ cos(θ) - r (θ) sin(θ))

dy/dx = (-cos(θ) sin(θ) + (1 - sin(θ)) cos(θ)) / (-cos²(θ) - (1 - sin(θ)) sin(θ))

dy/dx = - (cos(θ) - sin(2θ)) / (sin(θ) + cos(2θ))

The tangent is horizontal if dy/dx = 0 (or when the numerator vanishes):

cos(θ) - sin(2θ) = 0

cos(θ) - 2 sin(θ) cos(θ) = 0

cos(θ) (1 - 2 sin(θ)) = 0

cos(θ) = 0   or   1 - 2 sin(θ) = 0

cos(θ) = 0   or   sin(θ) = 1/2

[θ = π/2 + 2nπ   or   θ = 3π/2 + 2nπ]   or   [θ = π/6 + 2nπ   or   θ = 5π/6 + 2nπ]

where n is any integer.

In the interval 0 ≤ θ ≤ 2π, we get solutions of θ = π/6, θ = 5π/6, and θ = 3π/2. (We omit π/2 because the denominator is zero at that point and makes dy/dx undefined.) So the points where the tangent is horizontal are themselves (√3/4, 1/4), (-√3/4, 1/4), and (0, -2), respectively.

The tangent is vertical if 1/(dy/dx) = 0 (or when the denominator vanishes):

sin(θ) + cos(2θ) = 0

sin(θ) + (1 - 2 sin²(θ)) = 0

2 sin²(θ) - sin(θ) - 1 = 0

(2 sin(θ) + 1) (sin(θ) - 1) = 0

2 sin(θ) + 1 = 0   or   sin(θ) - 1 = 0

sin(θ) = -1/2   or   sin(θ) = 1

[θ = 7π/6 + 2nπ   or   θ = 11π/6 + 2nπ]   or   [θ = π/2 + 2nπ]

Then for 0 ≤ θ ≤ 2π, the tangent will be vertical for θ = 7π/6 and θ = 11π/6, which correspond respectively to the points (-3√3/4, -3/4) and (3√3/4, -3/4). (Again, we omit π/2 because this makes dy/dx non-existent.)

answer: what do you mean