Double angle identity. i really need some with it, but just 2 questions.

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

Step-by-step 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