INVERSE TRIGONOMETRIC FUNCTIONS 2
INTRODUCTION In chapter 1, we learnt that only one-one and onto functions are invertible. If a function f is one-one and onto, then its inverse exists and is denoted by f –1. We shall notice that trigonometric functions are not one-one over their whole domains and hence their inverse do not exist. It often happens that a given function f may not be one-one in the whole of its domain, but when we restrict it to a part of the domain, it may be so. If a function is one-one on a part of its domain, it is said to be inversible on that part only. If a function is inversible in serveral parts of its domain, it is said to have an inverse in each of these parts. In this chapter, we shall notice that trigonometric functions are not one-one over their whole domains and hence we shall restrict their domains to ensure the existence of their inverses and observe their behaviour through graphical representation. The inverse trigonometric functions play a very important role in calculus and are used extensively in science and engineering. REMARK → Let a real function f : Df Rf , where Df = domain of f and Rf = range of f, be invertible, –1 → –1 ∈ ∈ then f : Rf Df is given by f (y) = x iff y = f (x) for all x Df and y Rf . Thus, the roles of x and y are just interchanged during the transition from f to f –1 ∈ In fact, graph of f ={(x, y) : y = f (x), for all x Df } and –1 –1 ∈ graph of f ={(y, x) : x = f (y), for all y Rf } ∈ ={(y, x) : y = f (x), for all x Df }. Thus, (x, y) ∈ graph of f iff (y, x) ∈ graph of f –1. The point (y, x) is the reflection of the point (x, y) in the line y = x, therefore, the graph of f –1 can be obtained from the graph of f by reflecting it through the line y = x. 2.1 INVERSE TRIGONOMETRIC FUNCTIONS 1. Inverse sine function
Consider the sine function f defined by f (x) = sin x, Df (domain of f) = R and Rf (range of f ) = [– 1, 1]. Table for the graph of f :
5π 2π π π π π π π 2π 5π x … – π – – – – – 0 π … 6 3 2 3 6 6 3 2 3 6
1 3 3 1 1 3 3 1 y … 0 – – –1 – – 0 1 0 … 2 2 2 2 2 2 2 2 64 MATHEMATICS – XII A portion of the graph is shown in fig. 2.1.
Y y = 1 y = 1 2
π π π π 2π 5π 3 3π – π − O π π − 2 3 2 3 6 X 2 6 2
Fig. 2.1.
1 Note that the horizontal line y = meets its graph in many points, so f is not one-one. 2 π π But if we restrict the domain from – to both inclusive, we observe that in this part 2 2 ⎡ ππ⎤ of the domain f is one-one. Therefore, the function y = f (x) = sin x with Df = − , ⎣⎢ 22⎦⎥ and Rf = [– 1, 1] has an inverse function called the inverse sine function or the arc sine function, denoted by sin–1. Thus, Y π ππ 2 y = sin–1 x iff x = sin y and y ∈∈∈ ⎡– , ⎤ . ⎣⎢ 22⎦⎥ –1 Domain of sin x is [– 1, 1] –1 O 1 ππ X and its range is ⎡– , ⎤ . ⎣⎢ 22⎦⎥ –1 π The graph of sin x is shown in fig. 2.2. Note that it is − 2 one-one function. Fig. 2.2. REMARKS ππ (a) Besides ⎡– , ⎤ , there exist other intervals where the sine function is one-one and ⎣⎢ 22⎦⎥ hence has an inverse function but here by sin–1 x, we shall always mean the function ππ sin–1 : [– 1, 1] → ⎡– , ⎤ ⎣⎢ 22⎦⎥ π π defined above (unless stated otherwise). The portion of the curve for which – ≤ y ≤ 2 2 is known as the principal value branch of the function y = sin–1 x and these values of y are known as the principal values of the function y = sin–1 x. (b) The graph of sin–1 function can be obtained from the graph of the original
function by interchanging the roles of x x π –1 and y i.e. if (a, b) is a point on the graph of 2 sine function, then (b, a) becomes the 1 y = sin = sin corresponding point on the graph of y x –1 inverse sine function. The graph of sin π O π − –1 1 X function is the mirror image along the line 2 2 –1 y = x of the corresponding original π − function. This can be visualised by looking 2 the graphs of y = sin x and y = sin–1 x in the same axes as shown in fig. 2.3. Fig. 2.3. INVERSE TRIGONOMETRIC FUNCTIONS 65
⎡ ππ⎤ ⎡ ππ3 ⎤ (c) It may be noted that besides ⎢− , ⎥ , there exist other intervals such as ⎢ , ⎥, ⎣ 22⎦ ⎣ 2 2 ⎦ ⎡ 3ππ5 ⎤ ⎡ 3ππ⎤ ⎢ , ⎥, ⎢−−, ⎥ etc. where the sine function is one-one and hence has an inverse ⎣ 2 2 ⎦ ⎣ 22⎦ ⎡ ππ3 ⎤ ⎡ 3 ππ5 ⎤ ⎡ 3 π π⎤ function. Thus, the range of other branches are ⎢ ,,,,⎥ ⎢ ⎥ ⎢−− ,⎥ etc. ⎣ 2 2 ⎦ ⎣ 2 2 ⎦ ⎣ 22⎦ 2. Inverse cosine function Consider the cosine function f defined by
f (x) = cos x, Df = R and Rf = [– 1, 1]. A portion of the graph of cos x is shown in fig. 2.4.
Y 1
π O π – π − π X 2 2
–1
Fig. 2.4. Clearly, f is not one-one but if we restrict the domain to [0, π], f is one-one and so it has an inverse function called the inverse cosine function or the arc cosine function, denoted by cos–1. Thus, Y y = cos–1 x iff x = cos y and y ∈∈∈ [0, πππ ]. π Domain of cos–1 x is [– 1, 1] and its range is [0, π].
–1 The graph of cos x is shown in fig. 2.5. Note that it π is one-one function. 2 The portion of the curve for which 0 ≤ y ≤ π is known as the principal value branch of the function y = cos–1 x and these values of y are known as the principal values of the function y = cos–1 x. –1 O 1 X Note that the range of other branches of cos–1 x are [π, 2π], [2π, 3π], [– π, 0] etc. Fig. 2.5.
3. Inverse tangent function Consider the tangent function f defined by π f (x) = tan x, Df = R except odd multiples of Y 2 and Rf = R. A portion of the graph of tan x is shown in fig. 2.6. Clearly, f is not one-one but if we restrict the π π π ⎛ ππ⎞ – π − O X domain to – , , f is one-one and so it has 2 2 ⎝ 22⎠ an inverse function called the inverse tangent function or the arc tangent function, denoted by tan–1. Thus, ππ y = tan–1 x iff x = tan y and y ∈∈∈ ⎛ –,⎞ . ⎝ 22⎠ Fig. 2.6. 66 MATHEMATICS – XII
ππ Y Domain of tan–1 x is R and its range is ⎛ – , ⎞ . ⎝ ⎠ 22 π A portion of the graph of tan–1 x is shown in fig. 2 2.7. Note that it is one-one function. π π The portion of the curve for which – < y < O X 2 2 is known as the principal value branch of the –1 π function y = tan x and these values of y are − 2 known as the principal values.
Fig. 2.7.
4. Inverse cotangent function Y Consider the cotangent function f defined by π f (x) = cot x, Df = R except even multiples of and 2 Rf = R.
A portion of the graph of cot x is shown in fig. 2.8. – π O π π π − X 2 2 Clearly, f is not one-one but if we restrict the domain to (0, π ), f is one-one and so it has an inverse function called inverse cotangent function or arc cotangent function, denoted by cot –1. Thus,
y = cot –1 x iff x = cot y and y ∈∈∈ (0, πππ ). Fig. 2.8.
Y Domain of cot –1 x is R and its range is (0, π). π A portion of the graph of cot –1 x is shown in fig. 2.9. Note that it is one-one function. π The portion of the curve for which 0 < y < π is 2 known as the principal value branch of the O function y = cot –1 x and these values of y are X known as the principal values. Fig. 2.9. 5. Inverse secant function Consider the function f defined by f (x) = sec x,
π D = R except odd multiples of and f 2 Y Rf = (– ∞, – 1] ∪ [1, ∞). A portion of the graph of sec x is shown in fig. 2.10. Clearly, f is not one-one but if we restrict the 1 ⎡ π⎞ ⎛ π ⎤ domain to 0, ∪ , π , f is one-one and so π π 3π ⎣⎢ 2⎠ ⎝ 2 ⎦⎥ − O π X 2 –1 2 2 it has an inverse function called inverse secant function or arc secant function, denoted by sec–1. Thus,
–1 ⎡ π⎞ ⎛ π ⎤ y = sec x iff x = sec y and y ∈∈∈ 0, ∪ , π . Fig. 2.10. ⎣⎢ 2⎠ ⎝ 2 ⎦⎥ INVERSE TRIGONOMETRIC FUNCTIONS 67
π π Domain of sec–1 x is (– ∞ , – 1] ∪ [1, ∞) and its range is ⎡0, ⎞ ∪ ⎛ , π⎤. ⎣⎢ 2⎠ ⎝ 2 ⎦⎥ A portion of the graph of sec–1 x is shown in Y fig. 2.11. Note that it is one-one function. The portion of the curve for which 0 ≤ y ≤ π , π y ≠ , is known as the principal value branch 2 –1 of the function y = sec x and these values of y π are called the principal values. 2
–1 O 1 X 6. Inverse cosecant function Consider the function f defined by Fig. 2.11. f (x) = cosec x, D = R except even multiples of f Y π and Rf = (– ∞, – 1] ∪ [1, ∞). 2 A portion of the graph of cosec x is shown in fig. 2.12.
Clearly, f is not one-one but if we restrict the π 1 − 2 ⎡ π ⎞ ⎛ π⎤ domain to –,0 ∪ 0, , f is one-one and so – π O π ⎣⎢ 2 ⎠ ⎝ 2⎦⎥ π X –1 2 it has an inverse function called inverse cosecant function or arc cosecant function, denoted by cosec–1. Thus, y = cosec–1 x iff x = cosec y, π π y ∈∈∈ ⎡–,0⎞ ∪ ⎛0, ⎤. Fig. 2.12. ⎣⎢ 2 ⎠ ⎝ 2⎦⎥ Domain of cosec–1 x is (– ∞, –1] ∪ [1, ∞) and its range is Y ⎡ π ⎞ ⎛ π⎤ –,0 ∪ 0, . π ⎣⎢ 2 ⎠ ⎝ 2⎦⎥ 2 A portion of the graph of cosec–1 x shown in fig. 2.13. Note that it is one-one function. –1 O 1 The portion of the curve for which X π π π − – ≤ y ≤ , y ≠ 0, is known as the principal 2 2 2 value branch of the function y = cosec–1 x and these values of y are known as principal values. Fig. 2.13.
ILLUSTRATIVE EXAMPLES
Example 1. Find the principal values of : ⎛ 3 ⎞ ⎛ 3 ⎞ (i) cos–1 (ii) cosec–1 (2) (iii) sin–1 − . ⎝⎜ ⎠⎟ ⎜ ⎟ 2 ⎝ 2 ⎠ (NCERT)(NCERT)(C.B.S.E. 2010) ⎛ 3 ⎞ Solution. (i) Let cos–1 = x, 0 ≤ x ≤ π ⎝⎜ ⎠⎟ 2 3 3 π ⇒ = cos x ⇒ cos x = ⇒ cos x = cos 2 2 6 π ⎛ 3 ⎞ π ⇒ x = ⇒ cos–1 = . ⎝⎜ ⎠⎟ 6 2 6 68 MATHEMATICS – XII
π π (ii) Let cosec–1 (2) = x, – ≤ x ≤ , x ≠ 0 2 2 π ⇒ cosec x = 2 ⇒ cosec x = cosec 6 π π ⇒ x = ⇒ cosec–1 (2) = . 6 6 π π –1 − 3 ≤ ≤ (iii) Let sin = x, – x 2 2 2 π π ⇒ − 3 ⇒ ⇒ − sin x = sin x = – sin sin x = sin 2 3 3 π π ⇒ ⇒ –1 − 3 x = – sin = – . 3 2 3 Example 2. Find the principal values of : –1 − 1 –1 ()− –1 –1 (i) cos (ii) tan 3 (iii) cot (– 1) (iv) sec (– 2) 2 (NCERT) (NCERT) (C.B.S.E. 2010) –1 − 1 ≤ ≤ π Solution. (i) Let cos = x, 0 x 2 1 π ⇒ cos x = − ⇒ cos x = – cos 2 4 π π ⇒ π − 3 cos x = cos = cos 4 4 3π 1 3π ⇒ x = ⇒ cos–1 − = . 4 4 2 π π (ii) Let tan–1 ()− 3 = x, – < x < 2 2 π ⇒ − ⇒ tan x = 3 tan x = – tan 3 π π π ⇒ ⇒ ⇒ –1 ()− tan x = tan – x = – tan 3 = – . 3 3 3 (iii) Let cot–1 (– 1) = x, 0 < x < π π ⇒ cot x = – 1 ⇒ cot x = – cot 4 π π π π ⇒ π − 3 ⇒ 3 ⇒ –1 3 . cot x = cot = cot x = cot (– 1) = 4 4 4 4 π (iv) Let sec–1 (– 2) = x, 0 ≤ x ≤ π, x ≠ 2 π ⇒ sec x = – 2 ⇒ sec x = – sec 3 π π ⇒ π − ⇒ 2 sec x = sec sec x = sec 3 3
2π 2π ⇒ x = ⇒ sec–1 (– 2) = . 3 3 INVERSE TRIGONOMETRIC FUNCTIONS 69 Example 3. Find the principal values of : 3π 13π 7π 7π –1 –1 –1 –1 (i) sin sin (ii) cos cos (iii) cos cos (iv) tan tan . 5 6 6 6 (NCERT)(NCERT) (C.B.S.E. 2011, 09) (C.B.S.E. 2013) π π π –1 3 ≤ ≤ Solution. (i) Let sin sin = x, – x 5 2 2 π π π ⇒ 3 π − 2 2 sin x = sin = sin = sin 5 5 5 π π π ⇒ 2 ⇒ –1 3 2 x = sin sin = . 5 5 5 π –1 13 ≤ ≤ π (ii) Let cos cos = x, 0 x 6 π π π ⇒ 13 π + cos x = cos = cos 2 = cos 6 6 6 π π π ⇒ ⇒ –1 13 x = cos cos = . 6 6 6 π –1 7 ≤ ≤ π (iii) Let cos cos = x, 0 x 6 π π π ⇒ 7 π − 5 5 cos x = cos = cos 2 = cos 6 6 6 π ππ ⇒ 5 ⇒ –1 7 = 5 x = cos cos . 6 6 6 π π π –1 7 (iv) Let tan tan = x, – < x < 6 2 2 π π π ⇒ 7 π + tan x = tan = tan = tan 6 6 6 π π π ⇒ ⇒ –1 7 x = tan tan = . 6 6 6 Example 4. Using principal values, find the values of : –1 –1 –1 –1 − 1 –1 − 1 (i) tan 3 – sec (– 2) (ii) tan (1) + cos + sin . 2 2 (C.B.S.E. 2012)(NCERT) π π –1 Solution. (i) Let tan 3 = x, – < x < 2 2 π π ⇒ ⇒ tan x = 3 = tan x = . 3 3 π Let sec–1 (– 2) = y, 0 ≤ y ≤ π, y ≠ 2 π π π π ⇒ π 2 ⇒ 2 sec y = – 2 = – sec = sec – = sec y = . 3 3 3 3 ππ π ∴ –1 –1 −=−2 tan 3 – sec (– 2) = x – y = . 3 33 π π π π (ii) Let tan–1 (1) = x, – < x < ⇒ tan x = 1 = tan ⇒ x = . 2 2 4 4 –1 − 1 ≤ ≤ π Let cos = y, 0 y 2 π π π π ⇒ − 1 π 2 ⇒ 2 cos y = = – cos = cos – = cos y = . 2 3 3 3 3 70 MATHEMATICS – XII
π π –1 − 1 ≤ ≤ Let sin = z, – z 2 2 2 π π π ⇒ − 1 ⇒ sin z = = – sin = sin – z = – . 2 6 6 6 πππ ∴ –1 –1 − 1 –1 − 1 +−2 tan (1) + cos + sin = x + y + z = 2 2 4 36 (382 +− )π 9 3 = ==ππ. 12 12 4 π − 1 Example 5. Evaluate : sin −−sin 1 .(C.B.S.E. 2011, 08) 3 2
π π –1 − 1 ≤ ≤ Solution. Let sin = x, – x 2 2 2 π π ⇒ 1 − sin x = – = – sin = sin 2 6 6 π π ⇒ ⇒ –1 −=−1 x = – sin . 6 26 π − 1 ππ ππ π ∴ sin −−sin 1 = sin −− = sin + = sin = 1. 3 2 36 36 2
–1 4 Example 6. (i) If tan = x, find the value of cos x. 3 –1 − 1 (ii) If cot = x, find the values of sin x and cos x. 5 π π –1 4 ⇒ 4 Solution. (i) Given tan = x tan x = , – < x < . 3 32 2 For this value of x, sec x is +ve.
4 2 16 5 3 ∴ sec x = 11+=+tan2 x =+1 = ⇒ cos x = . 5 3 9 3 – 1 − 1 ⇒ 1 π (ii) Given cot = x cot x = – , 0 < x < . 5 5 For this value of x, cosec x is +ve.
2 ∴ +=+2 1 =+1 =26 ⇒ 5 cosec x = 11cot x – 1 sin x = . 5 25 5 26 cos x 1 5 1 Also cos x = . sin x = cot x . sin x = – . = – . sin x 5 26 26
Example 7. Evaluate the following : 3 5 (i) sin 2 cos–1 − (ii) sin 2cot−1 − (NCERT Examplar Problems) 5 12 1 5 (iii) tan cos–1 . 2 3
3 –1 − 3 ≤ ≤ π ⇒ ≤ ≤ π Solution. (i) Let cos = x, 0 x cos x = – , 0 x . 5 5 For this value of x, sin x ≥ 0, 2 ∴ 2 =−− 3 − 9 ==16 4 sin x = 11– cos x = 1 . 5 25 25 5 ∴ –1 − 3 4 − 3 =−24 sin 2 cos = sin 2x = 2 sin x cos x = 2. . . 5 5 5 25 INVERSE TRIGONOMETRIC FUNCTIONS 71
–1 − 5 π ⇒ − 5 π (ii) Let cot = x, 0 < x < cot x = , 0 < x < . 12 12 For this value of x, cosec x > 0.
2 ∴ +=+2 5 = 13 cosec x = 11cot x – 12 12 12 ⇒ sin x = . 13 cos x 5 12 =− 5 Now cos x = . sin x = cot x . sin x = – . . sin x 12 13 13 5 ∴ −1− sin 2 cot = sin 2x = 2 sin x cos x 12 12 −=−5 120 = 2. . . 13 13 169 1 5 5 π (iii) Let cos−1 = x ⇒ cos−1 = 2x, 0 ≤ 2x ≤ π i.e. 0 ≤ x ≤ 2 3 3 2
π−2 ⇒ 5 ≤≤ ⇒1 tan x = 5 cos 2x = , 0 x 2 321 + tan x 3 ⇒ 55+ tan 2 x = 3 – 3 tan2 x ⇒ ()35+ tan 2 x = 3 – 5
35− 35− π ⇒ tan 2 x = ⇒=tan x ( Q for 0 ≤ x ≤ , tan x ≥ 0) + + 2 35 35 − − − ⇒ 35× 35= 35 tan x = + − 3535 2 1 5 35− ⇒ tan cos–1 = . 2 3 2
EXERCISE 2.1 Very short answer type questions (1 to 21) : Find the principal values of the following (1 to 6) : 1 1 1. (i) sin–1 (NCERT)(ii) cos–1 (iii) tan–1 ()3 . 2 2 –1 2 –1 () –1 () 2. (i) sec (NCERT)(ii) cot 3 (NCERT)(iii) cosec 2 (NCERT) 3 –1 − 1 –1 –1 ()− 3. (i) cos (ii) tan (– 1) (iii) cosec 2 2 (C.B.S.E. 2011, 08 C)(NCERT) –1 − 1 –1 − 2 –1 . 4. (i) cot (NCERT)(ii) sec (iii) sin (– 1) 3 3 π 2π 3π –1 4 –1 –1 5. (i) cos cos (ii) sin sin (iii) tan tan 3 3 4 (NCERT)(C.B.S.E. 2011) π 9π 13π –1 5 –1 tan –1 cosec . 6. (i) cos cos (ii) tan 8 (iii) cosec 4 3 (C.B.S.E. 2013) 72 MATHEMATICS – XII 7. Show that : π π –1 5 ≠ 5 (i) tan tan , what is its value? 6 6 π π –1 − ≠− (ii) cos cos , what is its value? 66 π π –1 5 ≠ 5 (iii) sin sin . What is its value? 3 3 8. Using principal values, evalaute the following : π π –1 2 –1 2 cos cos + sin sin .(C.B.S.E. 2011, 08) 3 3 –1 9. Find the value of sec (tan (– 3 )). 10. Find the domain of the following functions : (i) sin x + sin–1 x (ii) sin –1 (1 – x). 11. Find the domain of the function cos –1 (3x – 2). 12. Find the domain of the function sec –1 (2x – 3). 13. Write the range of one branch of sin –1 x, other than the principal branch. (Sample Paper) 14. Write the range of one branch of cos –1 x, other than the principal branch. –1 3 15. If cot = x, find the value of cos x. 4 –1 3 16. (i) If tan = x, find the values of cos x and sin x. 4 –1 1 (ii) If cot – = x, find the values of sin x and cos x. 7 –1 –1 1 (iii) If tan x = sin , find the value of x. 2 17. Evaluate the following : π − 3 (i) cot (tan–1 3 )(ii) sin − sin 1 − 6 2
− 3 π (iii) cos cos 1 − + (NCERT Examplar Problems) 26
18. Prove the following : 1 − 3 2π 1 π (i) sin–1 − + cos 1 − = (ii) tan–1 (– 1) + cos–1 − = 2 2 3 2 2
19. Using principal values, find the values of : –1 1 –1 − 1 –1 –1 (i) cos – 2 sin (ii) tan ( 3 ) – cot (– 3 ) 2 2 (C.B.S.E. 2012)(C.B.S.E. 2013) 1 1 –1 –1 − –1 –1 − (iii) tan (1) + cos (iv) cosec (– 1) + cot . 2 3 (C.B.S.E. 2013) 20. Find the values of the following : − 1 − 3 (i) tan –1 22cos sin 1 (ii) tan–1 2sin 2 cos 1 2 2
(C.B.S.E. Sample Paper)(C.B.S.E. 2013) INVERSE TRIGONOMETRIC FUNCTIONS 73 21. Find the value of : ⎛ 5π⎞ ⎛ 13π⎞ (i) tan–1 ⎜tan ⎟ + cos–1 ⎜cos ⎟ (NCERT Examplar Problems) ⎝ 6 ⎠ ⎝ 6 ⎠ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ ⎛ π⎞⎞ (ii) tan–1 − + cot–1 + tan–1 sin⎜− ⎟ (NCERT Examplar Problems) ⎜ ⎟ ⎜ ⎟ ⎜ ⎝ ⎠⎟ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 2 ⎠ 22. Evaluate the following :
⎛ −1 ⎛ 3 ⎞⎞ ⎛ −1⎛ 12 ⎞⎞ ⎛ −1 3⎞ (i) cos sin – (ii) cosec cos − (iii) sin ⎜2 sin ⎟ . ⎝ ⎝ 5 ⎠⎠ ⎝ ⎝ 13 ⎠⎠ ⎝ 5⎠ (C.B.S.E. 2013) 2.2 PROPERTIES OF INVERSE TRIGONOMETRIC FUNCTIONS In this section, we shall prove some important properties of inverse trigonometric functions. These results are valid within the principal value branches of the corresponding inverse trigonometric functions and wherever they are defined. Some results may not be valid for all values of the domains of inverse trigonometric functions. In fact, they will be valid only for some values of x for which inverse trigonometric functions are defined. However, we may not go into the details of these values of x in the domain. 1. (i) sin (sin –1 x) = x, |x| ≤≤≤ 1 (ii) cos (cos –1 x) = x, |x| ≤≤≤ 1 (iii) tan (tan –1 x) = x, x ∈∈∈ R (iv) cot (cot –1 x) = x, x ∈∈∈ R (v) sec (sec –1 x) = x, |x| ≥≥≥ 1 (vi) cosec (cosec –1 x) = x, |x| ≥≥≥ 1. Proof. (i) Let sin –1 x = y ⇒ x = sin y ⇒ x = sin (sin –1 x)( Q y = sin –1 x) Thus, sin (sin –1 x) = x for all x in [– 1, 1] i.e. |x| ≤ 1. We leave the proofs of other parts for the reader.
–1 ⎡⎡ ππππ ⎤⎤ –1 2. (i) sin (sin x) = x, x ∈∈∈ ⎢⎢–,⎥⎥ (ii) cos (cos x) = x, x ∈∈∈ [0, πππ ] ⎣⎣ 22⎦⎦ ⎛⎛ ππππ⎞⎞ (iii) tan –1 (tan x) = x, x ∈∈∈ ⎜⎜–,⎟⎟ (iv) cot –1 (cot x) = x, x ∈∈∈ (0, πππ ) ⎝⎝ 22⎠⎠ –1 ⎡⎡ ππ⎞⎞ ⎛⎛ ππ ⎤⎤ (v) sec (sec x) = x, x ∈∈∈ ⎢⎢0, ⎟⎟ ∪∪∪ ⎜⎜ , ππ⎥⎥ ⎣⎣ 2 ⎠⎠ ⎝⎝ 2 ⎦⎦ –1 ⎡⎡ ππ ⎞⎞ ⎛⎛ ππ ⎤⎤ (vi) cosec (cosec x) = x, x ∈∈∈ ⎢⎢– , 0⎟⎟ ∪∪∪ ⎜⎜0, ⎥⎥ . ⎣⎣ 2 ⎠⎠ ⎝⎝ 2 ⎦⎦ Proof. (i) Let sin x = y ⇒ x = sin–1 y ⇒ x = sin–1 (sin x)( Q y = sin x)
–1 ⎡ ππ⎤ Thus, sin (sin x) = x for all x ∈ ⎢–,⎥ . ⎣ 22⎦ ⎡ ππ⎤ (It may be noted that when x ∈ ⎢–,⎥ , sin x is one-one and hence its inverse ⎣ 22⎦ function exists.) We leave the proofs of other parts for the reader. 3. (i) sin –1 (– x) = – sin –1 x, |x| ≤≤≤ 1 (ii) cos –1 (– x) = πππ – cos –1 x, |x| ≤≤≤ 1 (iii) tan –1 (– x) = – tan –1 x, x ∈∈∈ R (iv) cot –1 (– x) = πππ – cot –1 x, x ∈∈∈ R (v) cosec –1 (– x) = – cosec –1 x, |x| ≥≥≥ 1 (vi) sec –1 (– x) = πππ – sec –1 x, |x| ≥≥≥ 1. ππ Proof. (i) Let sin–1 x = y ⇒ x = sin y, – ≤≤y 22 ππ ⇒ – x = – sin y, ≥−y ≥− 22 ππ ⇒ – x = sin (– y), – ≤−y ≤ 22 ⇒ sin–1 (– x) = – y ⇒ sin–1 (– x) = – sin–1 x, | x | ≤ 1. (ii) Let cos –1 x = y ⇒ x = cos y, 0 ≤ y ≤ π INVERSE TRIGONOMETRIC FUNCTIONS 107 ANSWERS EXERCISE 2.1 π π π π π π 1. (i) (ii) (iii) 2. (i) (ii) (iii) 4 3 3 6 6 4 2π π π 2π 5π π 3. (i) (ii)– (iii) – 4. (i) (ii) (iii) – 3 4 4 3 6 2 2π π π π π π 5. (i) (ii) (iii) – 6. (i) (ii) (iii) – 3 3 4 3 8 4 π π π 7. (i) – (ii) (iii) – 8. π 9. 2 6 6 3 1 ππ3 10. (i)[– 1, 1] (ii) [0, 2] 11. 1, 12. (– ∞, 1] ∪ [2, ∞) 13. , 3 2 2 3 4 3 7 1 1 14. [π, 2π] 15. 16. (i) ; (ii) ; − (iii) 5 5 5 52 52 3 1 2π π 11π π 17. (i) (ii)1 (iii) – 1 19. (i) (ii) – (iii) (iv) 3 3 2 12 6 π π π 4 13 24 20. (i) (ii) 21. (i) 0 (ii) – 22. (i) (ii) (iii) 4 3 12 5 5 25 EXERCISE 2.2 π x –1 –1 4. (i) (ii) – x (iii) 2 tan x (iv) tan x 2 4 1 (v) cos–1 x (vi) 2 cos–1 x (vii) 2 sin–1 x (viii) 2 cos–1 x 2 π x –1 1 –1 –1 –1 x (ix) sin x (x) + tan x (xi) sin a (xii) 3 tan 4 2 a xy + 19 33 6 37 7. 10. (i) (ii) (iii) (iv) 1 − xy 8 65 17 26
1 1 45 2 12. (i) 3, – 3(ii) (iii) (iv) ± (v) ± 3 4 9 3 1 ab + (vi) 17 (vii) – (viii) − 12 1 ab 16. (i) 1 (ii) 0 17. (i) 1 (ii)0 π 4 15 x 18. (i) – (ii) (iii) 19. (i) (ii) sin–1 x 4 5 8 2 3 1 3 20. (i) 1 (ii) ± (iii) (iv) (v) 1 2 2 2 π π 2π π 21. (i) (ii) 22. (i) (ii) 4 4 5 2 EXERCISE 2.3
π 4 5 1 2. 3. 4. (i) (ii) – (iii)2 (iv) ab 3 3 3 2 π 3 5. nπ, nπ + , where n is any integer 6. x = , y = 0 4 2 7. x = 1, y = 2; x = 2, y = 7 CHAPTER TEST
24 π π 1 2. – 7. 8. 9. 0, ± 5 4 4 2