question 1 (1 point)\nthe interval(s) of increase for the sine function in the interval from 0° to 360° is…

question 1 (1 point)\nthe interval(s) of increase for the sine function in the interval from 0° to 360° is (are)\na) {x ∈ ℝ, 90° ≤ x ≤ 270°}\nb) {x ∈ ℝ, 0° ≤ x ≤ 180°}\nc) {x ∈ ℝ, 180° ≤ x ≤ 360°}\nd) {x ∈ ℝ, 0° ≤ x ≤ 90°, 270° ≤ x ≤ 360°}

question 1 (1 point)\nthe interval(s) of increase for the sine function in the interval from 0° to 360° is (are)\na) {x ∈ ℝ, 90° ≤ x ≤ 270°}\nb) {x ∈ ℝ, 0° ≤ x ≤ 180°}\nc) {x ∈ ℝ, 180° ≤ x ≤ 360°}\nd) {x ∈ ℝ, 0° ≤ x ≤ 90°, 270° ≤ x ≤ 360°}

Answer

Explanation:

Step1: Recall sine - function properties

The sine function (y = \sin(x)) has a derivative (y'=\cos(x)). The function is increasing when (y'>0), i.e., (\cos(x)>0).

Step2: Find intervals in (0^{\circ}\leq x\leq360^{\circ})

In the interval (0^{\circ}\leq x\leq360^{\circ}), (\cos(x)>0) when (0^{\circ}\leq x < 90^{\circ}) and (270^{\circ}<x\leq360^{\circ}). So the sine - function (y = \sin(x)) is increasing in the intervals ({x\in\mathbb{R},0^{\circ}\leq x\leq90^{\circ},270^{\circ}\leq x\leq360^{\circ}}).

Answer:

D. ({x\in\mathbb{R},0^{\circ}\leq x\leq90^{\circ},270^{\circ}\leq x\leq360^{\circ}})