Question:
Let θ = arccos (-1/2), then θ = -4π/3 and cosθ = -1/2
Last updated: 8/4/2022
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Let θ = arccos (-1/2), then θ = -4π/3 and cosθ = -1/2 θ = 2π/3 and cosθ = -1/2 θ = 4π/3 and cosθ = -(1/2) θ = -(2π/3) and cosθ = -1/2
Last updated: 8/4/2022
Let θ = arccos (-1/2), then θ = -4π/3 and cosθ = -1/2 θ = 2π/3 and cosθ = -1/2 θ = 4π/3 and cosθ = -(1/2) θ = -(2π/3) and cosθ = -1/2