Section 10.6: The Inverse Trigonometric Functions - Ostts.org

834 Foundations of Trigonometry(c) Since the argument −2 is negative, we cannot directly apply Theorem 10.27 to help usfind arccot(−2). Let t = arccot(−2). Then t is a real number such that 0 < t < πand cot(t) = −2. Moreover, since cot(t) < 0, we know π 2< t < π. Geometrically, thismeans t corresponds to a Quadrant II angle θ = t radians. This allows us to proceedusing a ‘reference angle’ approach. Consider α, the reference angle for θ, as picturedbelow. By definition, α is an acute angle so 0 < α < π 2, and the Reference AngleTheorem, Theorem 10.2, tells us that cot(α) = 2. This means α = arccot(2) radians.Since the argument of arccotangent is now a positive 2, we can use Theorem 10.27 to getα = arccot(2) = arctan ( (12)radians. Since θ = π − α = π − arctan12)≈ 2.6779 radians,we get arccot(−2) ≈ 2.6779.y1θ = arccot(−2) radiansα1xAnother way to attack the problem is to use arctan ( − 1 2). By definition, the real numbert = arctan ( − 2) 1 satisfies tan(t) = −12 with − π 2 < t < π 2. Since tan(t) < 0, we knowmore specifically that − π 2< t < 0, so t corresponds to an angle β in Quadrant IV. Tofind the value of arccot(−2), we once again visualize the angle θ = arccot(−2) radiansand note that it is a Quadrant II angle with tan(θ) = − 1 2. This means it is exactly πunits away from β, and we get θ = π + β = π + arctan ( − 1 2)≈ 2.6779 radians. Hence,as before, arccot(−2) ≈ 2.6779.