please trigonometry help

Question

x is in quadrant 3, approximate sin (2 x) if cos (x) = - 0.2. Round your answer to two decimal places.

Answer 1

Given tht cos(x) = -0.2

We have sin^2(x) + cos^2(x) = 1

                 sin^2(x) = 1 - cos^2(x)

                 sin(x)^2  = 1- (0.04)

                 sin(x)^2 = 0.96

                 sin(x) = +/- √(96 / 100)

                 sin(x) = - √(24 / 25)

                 sin(x) = -2 * √(6) / 5

We have to find the value of sin(2x)

Therefore sin(2x) = 2 * sin(x) * cos(x)                [ Since sin(2x) = 2 * sin(x) * cos(x) formula ]

                            = 2 * (-2 * sqrt(6) / 5) * (-1/5)

                            = (4 * sqrt(6)) / 25 

                            = 4* 2.4494/25

                            = 0.39191835884530849571156545195294

                            = 0.39

Therefore the approximate value of sin(2x) is 0.39.

Answer 2

The value of cos(x) = - 0.2 = - 2/10 = - 1/5 and the angle x lies in third quadrant.

Double angle formula : sin(2θ) = 2 sin(θ) * cos(θ).

Find the value of sin(x), by using Pythagorean theorem sin²(θ) + cos²(θ) = 1, to obtain

sin²(θ) + (-1/5)² = 1

sin²(θ) = 1 - 1/25 = (25-1)/25 = 24/25

sin(θ) = ± [√24/√25] = ± 2√6/5.

The sine function is negative in third quadrant, so negative root to obtain

sin(θ) = - 2√6/5

sin(2x) = 2 sin(x) * cos(x)

             = 2 *(- 2√6/5) * (- 1/5)

             = 4(√6)/25.

             = 0.16*√6.

             = 0.39.