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True . The angle, θ 1 θ 1 that equals arccos (− x) arccos (− x), x > 0 x > 0, will be a second quadrant angle with reference angle, θ 2 θ 2, where θ 2 θ 2 equals arccos x arccos x, x > 0 x > 0. Since θ 2 θ 2 is the reference angle for θ 1 θ 1, θ 2 = π − θ 1 θ 2 = π − θ 1 and arccos (− x) arccos (− x) = π − arccos ...

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