Answer:
a) sin(2θ) = -√3/2
cos(2θ) = -1/2
tan(2θ) = √3
b) sin(2θ) = 4√5/9
cos(2θ) = 1/9
tan(2θ) = 4√5
Explanation:
a) cos θ = -1/2
First, find sin θ using Pythagorean identity:
sin² θ + cos² θ = 1
sin² θ + (-1/2)² = 1
sin² θ + 1/4 = 1
sin² θ = 3/4
sin θ = ±√3/2
Since θ is in Quadrant II, sin θ > 0, so sin θ = +√3/2.
Now use double angle formulas to find each expression.
sin(2θ) = 2 sin θ cos θ
sin(2θ) = 2 (√3/2) (-1/2)
sin(2θ) = -√3/2
cos(2θ) = cos² θ − sin² θ
cos(2θ) = (-1/2)² − (√3/2)²
cos(2θ) = 1/4 − 3/4
cos(2θ) = -1/2
tan(2θ) = sin(2θ) / cos(2θ)
tan(2θ) = (-√3/2) / (-1/2)
tan(2θ) = √3
b) sin θ = -2/3
Repeat same steps as before.
First, find cos θ using Pythagorean identity:
sin² θ + cos² θ = 1
(-2/3)² + cos² θ = 1
4/9 + cos² θ = 1
cos² θ = 5/9
cos θ = ±√5/3
Since θ is in Quadrant III, cos θ < 0, so cos θ = -√5/3.
Now use double angle formulas to find each expression.
sin(2θ) = 2 sin θ cos θ
sin(2θ) = 2 (-2/3) (-√5/3)
sin(2θ) = 4√5/9
cos(2θ) = cos² θ − sin² θ
cos(2θ) = (-√5/3)² − (-2/3)²
cos(2θ) = 5/9 − 4/9
cos(2θ) = 1/9
tan(2θ) = sin(2θ) / cos(2θ)
tan(2θ) = (4√5/9) / (1/9)
tan(2θ) = 4√5