Section 10.6: The Inverse Trigonometric Functions - Ostts.org

10.6 The Inverse Trigonometric Functions 833We now see what these restrictions mean in terms of x. Since 4x = csc(t), we get thatfor 0 ≤ t ≤ π 2 , 4x ≥ 1, or x ≥ 1 4. In this case, we can simplify |x| = x socos(t) =√16x 2 − 14|x|=√16x 2 − 14xSimilarly, for π < t ≤ 3π 2 , we get 4x ≤ −1, or x ≤ − 1 4. In this case, |x| = −x, so we alsoget√ √ √16xcos(t) = −2 − 1 16x= −2 − 1 16x=2 − 14|x| 4(−x) 4x√16x 2 −14xHence, in all cases, cos(arccsc(4x)) = , and this equivalence is valid for all x inthe domain of t = arccsc(4x), namely ( −∞, − 1 ] [4 ∪ 14 , ∞)10.6.3 Calculators and the Inverse Circular Functions.In the sections to come, we will have need to approximate the values of the inverse circular functions.On most calculators, only the arcsine, arccosine and arctangent functions are available and theyare usually labeled as sin −1 , cos −1 and tan −1 , respectively. If we are asked to approximate thesevalues, it is a simple matter to punch up the appropriate decimal on the calculator. If we are askedfor an arccotangent, arcsecant or arccosecant, however, we often need to employ some ingenuity, asour next example illustrates.Example 10.6.5.1. Use a calculator to approximate the following values to four decimal places.(a) arccot(2) (b) arcsec(5) (c) arccot(−2) (d) arccsc(− 3 )22. Find the domain and range of the following functions. Check your answers using a calculator.(a) f(x) = π ( x)( x)2 − arccos (b) f(x) = 3 arctan (4x). (c) f(x) = arccot + π52Solution.1. (a) Since 2 > 0, we can use the property listed in Theorem 10.27 to rewrite arccot(2) asarccot(2) = arctan ( 12). In ‘radian’ mode, we find arccot(2) = arctan( 12)≈ 0.4636.(b) Since 5 ≥ 1, we can use the property from either Theorem 10.28 or Theorem 10.29 towrite arcsec(5) = arccos ( 15)≈ 1.3694.