Question

Find 2 values of theta in degrees (0 deg <_ theta <_ 360 deg) for the angle sin theta= sqrt2/2

Answer 1

in first and second quadrants.

 

is the answer.

Answer 2

The trigonometric equation is sin(θ) = √2/2, where 0^o ≤ θ ≤ 360^o.

sin(θ) = √2/2 = 1/√2.

The function sin(θ) has a period of 2π first find all solutions in the interval [0, 2π].

The function sin(θ) is positive in Quadrant I and Quadrant II.

In first quadrant, 0^o ≤ θ ≤ 90^o.

1/√2 = sin(θ)  -------> sin(θ) = sin(π/4) ---------> θ = π/4.

In second quadrant, 90^o ≤ θ ≤ 180^o.

1/√2 = sin(π/4) = sin(π - π/4) = sin(3π/4)  -------> sin(θ) = sin(3π/4) ---------> θ = 3π/4.

The values of θ are π/4 and 3π/4 in 0^o ≤ θ ≤ 360^o.