The given range for θ is -π/2 ≤ θ ≤ π/2. This range is in the fourth and first quadrants of the unit circle.

The sine function is negative in the fourth quadrant.

The reference angle for sin(θ) = 1/2 is θ = π/6 or 30 degrees.

However, since the sine function is negative, we are looking for the angle in the fourth quadrant.

In the fourth quadrant, the angle with a reference angle of π/6 is -π/6.

So, θ = -π/6.

Question

The given range for θ is -π/2 ≤ θ ≤ π/2. This range is in the fourth and first quadrants of the unit circle.

The sine function is negative in the fourth quadrant.

The reference angle for sin(θ) = 1/2 is θ = π/6 or 30 degrees.

However, since the sine function is negative, we are looking for the angle in the fourth quadrant.

In the fourth quadrant, the angle with a reference angle of π/6 is -π/6.

So, θ = -π/6.

Knowee AI · Accepted Answer

The given range for θ is -π/2 ≤ θ ≤ π/2. This range is in the fourth and first quadrants of the unit circle.

The sine function is negative in the fourth quadrant.

The reference angle for sin(θ) = 1/2 is θ = π/6 or 30 degrees.

However, since the sine function is negative, we are looking for the angle in the fourth quadrant.

In the fourth quadrant, the angle with a reference angle of π/6 is -π/6.

So, θ = -π/6.

–𝜋2≤θ≤𝜋2. Find the value of θ in radians.sin(θ)=–12Write your answer in simplified, rationalized form. Do not round.θ=