Question

Find exact values of sin2theta, cos2theta, and tan2theta. Given: cos(theta)=-2/3

Answer 1

Given that cosθ = -2/3

Let cos2θ = 2*(cosθ)^2 - 1                 [ Trignometric formula ]

       cos2θ = 2*(-2/3)^2 - 1

       cos2θ = 2*(4/9) - 1

       cos2θ = 8/9 - 1

       cos2θ = -1/9

Therefore  cos2θ = -1/9

Let sin^2 + cos^2 = 1                         [ Trignometric formula ]

      (sin2θ)^2 + (cos2θ)^2 = 1

      (sin2θ)^2 + (-1/9)^2 = 1

      (sin2θ)^2 + 1/81 = 1

      (sin2θ)^2 = 1 - 1/81

      (sin2θ)^2 = 80/81

      sin2θ = √(80/81)

      sin2θ = √((16X5)/81)

                 = (4√5)/9

Therefore   sin2θ= (4√5)/9

Let tan2θ = sin2θ/cos2θ           [ Since tanθ = sinθ/cosθ ]

       tan2θ = ((4√5)/9)/(-1/9)

       tan2θ = -4√5

Therefore tan2θ = -4√5

Therefore the exact values of sin2θ, cos2θ and tan2θ are (4√5)/9, -1/9 and -4√5 respectively.

Answer 2

The value of cos(θ) = - 2/3.

cos(θ) is negative in second and third quadrants.

 

In second quadrant :

Double angle formula : cos(2θ) = 2 cos²(θ) - 1.

cos(2θ) = 2 * (- 2/3)² - 1

               = 8/9 - 1

               = - 1/9.

Pythagorean identity : sin² θ + cos² θ = 1

sin² (2θ) + cos² (2θ) = 1

sin² (2θ) + (- 1/9)² = 1

sin² (2θ) = 1 - 1/81 = 80/81

sin(2θ) = ± √80/√81

sin(θ) is positive in second quadrant , so positive root obtain

sin(2θ) = 4√5/9.

Quotient  identity : tan(θ) = sin(θ)/cos(θ)

tan(2θ) = sin(2θ)/cos(2θ)

              = [4√5/9] / [- 1/9]

              = - 4√5.

 

In third quadrant :

The trigonometric functions sin(θ) is negative, tan(θ) is positive in third quadrant.

sin(2θ) = - 4√5/9 and tan(2θ) = 4√5.