Basic Concepts of Inverse Trigonometric Function

MATHS INVERSE TRIGONOMETRIC FUNCTIONS BASIC CONCEPTS OF INVERSE TRIGONOMETRIC FUNCTION INVERSE FUNCTION Definition If a function is one to one and onto from A to B, then function g which associates each element y  B to one and only one element x  A, such that y = f (x), then g is called the inverse function of f, denoted by x = g (y). Usually we denote g = f –1 {Read as f inverse}  x = f–1(y). Inverse Trigonometric Functions We have seen that the trigonometric functions, sin, cos etc. are all periodic and thus, each of them achieves the same numerical value at an infinite number of points. Thus, the equation sin x = 1 2 has an infinite number of solutions, viz., x = π 6 ,π − π 6 etc. If one is to answer 1 the question : “What is the angle whose sine is 2 ?”, there is no unique answer. The difficulty arises as the function f : R → R defined by f(x) = sin x is not one to one and thus, does not admit of an inverse. To achieve a unique answer to the aforesaid question, we restrict the domain and codomain of sinx so that the resulting function is invertible. Thus, the function g: − ,   →−1,1 defined by  2 2 g(x) = sin x is one to one and onto and admits of an inverse (denoted by h = sin-1and read as sine inverse or arc sin) defined as h:−1,1 →− ,    2 2 where h(y) = x if y = sin x. The function sin–1 is the inverse of the sine function when the sine function is viewed in a restricted sense. We similarly define the other inverse trigonometric functions 1 MATHS ntervals for Inverse Functions Here, sin–1x, cosec–1x, tan–1x belongs to I and IV quadrant. /2 I IV –/2 Here, cos–1x, sec–1x, cot–1x belongs to I and II quadrant. II I  O IV ` 1. I quadrant is common to all the inverse functions. 2. III quadrant is not used in inverse function. 3. IV quadrant is used in the clockwise direction i.e., –/2  y  0. Domain, Range and Graphs of Inverse Functions 1. If sin y = x, then y = sin–1x, under certain condition. –1  sin y  1; but sin y = x.  –1  x  1 Again, sin y = –1  y = – /2 And sin y = 1  y = /2 Keeping in mind numerically smallest angles or real numbers.  – /2  y  /2 2 MATHS These restrictions on the values of x and y provide us with the domain and range for the function y = sin–1x. i.e., Domain : x  [–1, 1] Range : y  [–/2, /2] y (1, /2) –1 O y = sin x x (–1, –/2) 2. Let cos y = x then y = cos–1x, under certain condition –1  cos y  1.  –1  x  1 cos y = –1  y =  cos y = 1  y = 0  0  y   {as cos x is a decreasing function in [0, ]; hence cos  cos y  cos 0} These restrictions on the values of x and y provide us the domain and range for the function y = cos–1x. i.e., Domain : x  [–1, 1] Range : y  [0, ] (–1,) y y = cos–1x x O (1, 0) 3. If tan y = x then y = tan–1x, under certain conditions. 3 MATHS y Here, tan y  R  x  R ; –  < tan y <   –/2 < y < /2 y = /2 Thus, domain x  R Range y  (–/2, /2) 4. If cot y = x, then y = cot–1x (under certain conditions) cot y  R  x  R; –  < cot y <   0 < y <  These conditions on x and y make the function, cot y = x one-one and onto so that the inverse function exists. i.e., y = cot–1x is meaningful. i.e., Domain : x  R Range : y  (0, ) y= (0, /2) x O 5. If sec y = x, then y = sec–1x, where | x |  1 and 0  y  , y  /2 Here, Domain : x  R – (–1, 1) Range : y  [0, ] – {/2} (–1, ) y = /2 O x O x (1, 0) –1 y = sec x 4 y = –/2 –1 y = tan x MATHS 6. If cosec y = x then y = cosec–1x, y where | x |  1 and –/2  y  /2, y  0 (1, /2) Here, domain  R – (–1, 1) x Range  [–/2, /2] – {0} (–1, –) Function sin–1x Domain [–1, 1] Range − ,   2 2 cos–1x [–1, 1] [0, ]   tan–1x (–, ) − ,   2 2 cot–1x (–, ) (0, )  sec–1x (–,–1] [1, ) 0,   ,   2  2  cosec–1 x (–,–1] [1, ) – , 0 0,   2   2 Ex.1:   1  1 Find the value of tan cos−1   +tan−1 −  . 2 Sol: tan cos−1   +tan−1 −   2  3        3   1  1        = tan  +−  3 6 1 .  = tan   = 6   3 Ex.2: Find domain of sin–1 (2x2 – 1) Sol.: Let y = sin–1 (2x2 – 1) For y to be defined – 1  (2x2 – 1)  1  Ex.3 0  2x2  2  0  x2  1  x  [–1, 1].  1  2  1  2 The value of tan–1(1) + cos–1 − + sin–1 −  is equal to5 O –1 y = cosec x MATHS Sol.  1  2  1  2 tan–1(1) + cot–1 − + sin–1 −  =  + 2 –  =  +  = 3 4 3 6 4 2 4 Ex.4 1 1 The value of cos  cos−1  is equal to- Sol. Let cos–1 = , 2 8 1 8 where 0 <  < . 2 Then 1 cos−1 1 = 1  2 8 2 1 1 1  cos  cos−1  = cos  2 8 Nowcos–1  cos = 2 1=  8 1 8  1  2cos2 – 1 = 2  cos2  cos [ 0 < 8 = 9 2 16 = 3 2 4  <  , so cos   – 3 ] 2 4 2 4 6