828 Foundations of Trigonometry<strong>The</strong>orem 10.28. Properties of the Arcsecant and Arccosecant <strong>Functions</strong> a• Properties of F (x) = arcsec(x)– Domain: {x : |x| ≥ 1} = (−∞, −1] ∪ [1, ∞)– Range: [ 0, π ) (2 ∪ π2 , π]– as x → −∞, arcsec(x) → π + 2; as x → ∞, arcsec(x) → π −2– arcsec(x) = t if and only if 0 ≤ t < π 2 or π 2< t ≤ π and sec(t) = x– arcsec(x) = arccos ( 1x)provided |x| ≥ 1– sec (arcsec(x)) = x provided |x| ≥ 1– arcsec(sec(x)) = x provided 0 ≤ x < π 2 or π 2 < x ≤ π• Properties of G(x) = arccsc(x)– Domain: {x : |x| ≥ 1} = (−∞, −1] ∪ [1, ∞)– Range: [ − π 2 , 0) ∪ ( 0, π ]2– as x → −∞, arccsc(x) → 0 − ; as x → ∞, arccsc(x) → 0 +– arccsc(x) = t if and only if − π 2 ≤ t < 0 or 0 < t ≤ π 2and csc(t) = x– arccsc(x) = arcsin ( 1x)provided |x| ≥ 1– csc (arccsc(x)) = x provided |x| ≥ 1– arccsc(csc(x)) = x provided − π 2 ≤ x < 0 or 0 < x ≤ π 2– additionally, arccosecant is odda . . . assuming the “Trigonometry Friendly” ranges are used.Example <strong>10.6</strong>.3.1. Find the exact values of the following.(a) arcsec(2) (b) arccsc(−2) (c) arcsec ( sec ( ))5π4(d) cot (arccsc (−3))2. Rewrite the following as algebraic expressions of x and state the domain on which the equivalenceis valid.(a) tan(arcsec(x))(b) cos(arccsc(4x))
<strong>10.6</strong> <strong>The</strong> <strong>Inverse</strong> <strong>Trigonometric</strong> <strong>Functions</strong> 829Solution.1. (a) Using <strong>The</strong>orem 10.28, we have arcsec(2) = arccos ( )12 =π3 .(b) Once again, <strong>The</strong>orem 10.28 comes to our aid giving arccsc(−2) = arcsin ( − 2) 1 = −π(c) Since 5π 4doesn’t fall between 0 and π 2 or π 2and π, we cannot use the inverse propertystated in <strong>The</strong>orem 10.28. We can, nevertheless, begin by working ‘inside out’ whichyields arcsec ( sec ( )) √ ( √ )5π4 = arcsec(− 2) = arccos − 22= 3π 4 .(d) One way to begin to simplify cot (arccsc (−3)) is to let t = arccsc(−3). <strong>The</strong>n, csc(t) = −3and, since this is negative, we have that t lies in the interval [ − π 2 , 0) . We are aftercot (arccsc (−3)) = cot(t), so we use the Pythagorean Identity 1 + cot 2 (t) = csc 2 (t).Substituting, we have 1 + cot 2 (t) = (−3) 2 , or cot(t) = ± √ 8 = ±2 √ 2. Since − π 2 ≤ t < 0,cot(t) < 0, so we get cot (arccsc (−3)) = −2 √ 2.2. (a) We begin simplifying tan(arcsec(x)) by letting t = arcsec(x). <strong>The</strong>n, sec(t) = x for t in[0,π2)∪( π2 , π] , and we seek a formula for tan(t). Since tan(t) is defined for all t valuesunder consideration, we have no additional restrictions on t. To relate sec(t) to tan(t), weuse the identity 1+tan 2 (t) = sec 2 (t). This is valid for all values of t under consideration,and when we substitute sec(t) = x, we get 1 + tan 2 (t) = x 2 . Hence, tan(t) = ± √ x 2 − 1.If t belongs to [ 0, π 2)then tan(t) ≥ 0; if, on the the other hand, t belongs to( π2 , π] thentan(t) ≤ 0. As a result, we get a piecewise defined function for tan(t)6 .tan(t) ={ √x 2 − 1, if 0 ≤ t < π 2− √ x 2 − 1,if π 2 < t ≤ πNow we need to determine what these conditions on t mean for x. Since x = sec(t),when 0 ≤ t < π 2 , x ≥ 1, and when π 2< t ≤ π, x ≤ −1. Since we encountered no furtherrestrictions on t, the equivalence below holds for all x in (−∞, −1] ∪ [1, ∞).tan(arcsec(x)) ={ √x 2 − 1, if x ≥ 1− √ x 2 − 1,if x ≤ −1(b)[To simplify cos(arccsc(4x)), we start by letting t = arccsc(4x). <strong>The</strong>n csc(t) = 4x for t in−π2 , 0) ∪ ( 0, π ]2 , and we now set about finding an expression for cos(arccsc(4x)) = cos(t).Since cos(t) is defined for all t, we do not encounter any additional restrictions on t.From csc(t) = 4x, we get sin(t) = 14x, so to find cos(t), we can make use if the identitycos 2 (t) + sin 2 (t) = 1. Substituting sin(t) = 14x gives cos2 (t) + ( 1√ √16xcos(t) = ±2 − 1 16x16x 2 = ±2 − 14|x|4x) 2 = 1. Solving, we getSince t belongs to [ − π 2 , 0) ∪ ( 0, π ] √2 , we know cos(t) ≥ 0, so we choose cos(t) = 16−x 24|x|.(<strong>The</strong> absolute values here are necessary, since x could be negative.) To find the values for
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