INVERSE TRIGONOMETRIC FUNCTIONS

Inverse Trigonometric MODULE - IV 8 INVERSE TRIGONOMETRIC FUNCTIONS In the previous lesson, you have studied the definition of a function and different kinds of functions. We have defined inverse function. Let us briefly recall : Let f be a one-one onto function from A to B. Let y be an arbitary element of B. Then, f being onto, an element A such that f = y. Also, f being one-one, then must be unique. Thus for each y B, a unique element A such that f = y. So we may define a function, denoted by f as f :B A f ( y) = f = y The above function f is called the inverse of f. A function is invertiable if and only if f is one-one onto. In this case the domain of f is the range of f and the range of f is the domain f. Let us take another eample. We define a function : f : Car Registration No. If we write, g : Registration No. Car, we see that the domain of f is range of g and the range of f is domain of g. So, we say g is an inverse function of f, i.e., g = f. In this lesson, we will learn more about inverse trigonometric function, its domain and range, and simplify epressions involving inverse trigonometric functions. OBJECTIVES After studying this lesson, you will be able to : define inverse trigonometric functions; MATHEMATICS 4

MODULE - IV Inverse Trigonometric state the condition for the inverse of trigonometric functions to eist; define the principal value of inverse trigonometric functions; find domain and range of inverse trigonometric functions; state the properties of inverse trigonometric functions; and simplify epressions involving inverse trigonometric functions. EXPECTED BACKGROUND KNOWLEDGE Knowledge of function and their types, domain and range of a function Formulae for trigonometric functions of sum, difference, multiple and sub-multiples of angles. 8. IS INVERSE OF EVERY FUNCTION POSSIBLE? Take two ordered pairs of a function (,y ) and (,y) If we invert them, we will get ( y, ) and ( y, ) This is not a function because the first member of the two ordered pairs is the same. Now let us take another function : sin,, sin, 4 and 3 sin, 3 Writing the inverse, we have,sin,,sin 4 and 3,sin 3 which is a function. Let us consider some eamples from daily life. f : Student Score in Mathematics Do you think f will eist? It may or may not be because the moment two students have the same score, f will cease to be a function. Because the first element in two or more ordered pairs will be the same. So we conclude that every function is not invertible. Eample 8. If f : R R defined by 3 f () = + 4. What will be f? Solution : In this case f is one-to-one and onto both. f is invertible. Let 3 y= +4 y 4 = = y 4 3 3 So f, inverse function of f i.e., f ( y) = 3 y4 4 MATHEMATICS

Inverse Trigonometric The functions that are one-to-one and onto will be invertible. Let us etend this to trigonometry : Take y = sin. Here domain is the set of all real numbers. Range is the set of all real numbers lying between and, including and i.e. y. We know that there is a unique value of y for each given number. In inverse process we wish to know a number corresponding to a particular value of the sine. Suppose y= sin = 5 3 sin = sin = sin = sin =... 6 6 6 5 3 may have the values as,, = 6 6 6 Thus there are infinite number of values of. y = sin can be represented as..., 6, 5,,... 6 The inverse relation will be,, 6 5,,... 6 MODULE - IV It is evident that it is not a function as first element of all the ordered pairs is, which contradicts the definition of a function. Consider y = sin, where R (domain) and y [, ] or y which is called range. This is many-to-one and onto function, therefore it is not invertible. Can y = sin be made invertible and how? Yes, if we restrict its domain in such a way that it becomes one-to-one and onto taking as (i) (ii) (iii), y [, ] or 3 5 y [, ] or 5 3 y [, ] etc. Now consider the inverse function y= sin. We know the domain and range of the function. We interchange domain and range for the inverse of the function. Therefore, MATHEMATICS 43

MODULE - IV (i) (ii) (iii) y [, ] or 3 5 y [, ] or 5 3 y [, ] etc. Inverse Trigonometric Here we take the least numerical value among all the values of the real number whose sine is which is called the principle value of sin. For this the only case is y. Therefore, for principal value of is [, ] i.e. [, ] and range is y. Similarly, we can discuss the other inverse trigonometric functions. Function Domain Range (Principal value). y= sin [, ],. y= [, ] [0, ] 3. y= tan R, 4. y= cot R [0, ] y= sin, the domain 5. y= sec or 0,, 6. y= ec or,0 0, 8. GRAPH OF INVERSE TRIGONOMETRIC FUNCTIONS 8. Gr y= sin y= 44 MATHEMATICS

Inverse Trigonometric MODULE - IV y= tan y= cot y= sec y= ec Fig. 8. Eample 8. Find the principal value of each of the following : (i) sin (ii) (iii) tan 3 Solution : (i) Let sin =θ or sinθ= = sin 4 or θ= 4 (ii) Let =θ θ= = = 3 3 or θ= 3 (iii) Let tan =θ 3 or = tanθ or tanθ= tan 3 6 θ= 6 MATHEMATICS 45

MODULE - IV Inverse Trigonometric Eample 8.3 Find the principal value of each of the following : (a) (i) (ii) tan (b) Find the value of the following using the principal value : sec 3 Solution : (a) (i) Let =θ, then = θ or θ= 4 θ= 4 (ii) Let tan ( ) =θ, then = tan θ or tanθ= tan 4 θ= 4 (b) Let 3 =θ, then 3 = θ or θ= 6 θ= 6 sec 3 sec sec = θ= = 6 3 Eample 8.4 Simplify the following : (i) [ ] (ii) [ cot ec ] sin Solution : (i) Let sin =θ = sinθ [ ] (ii) Let sin = θ= sin θ= ec =θ = ec θ 46 MATHEMATICS

Inverse Trigonometric Also cotθ= ec θ MODULE - IV = CHECK YOUR PROGRESS 8.. Find the principal value of each of the following : (a) 3 (d) tan ( 3) (b) ec ( ) (e) cot. Evaluate each of the following : (a) 3 (d) ( tan sec ) (e) ec cot ( 3 ) 3. Simplify each of the following epressions : (a) ( sec tan ) (d) ( cot ) (c) sin 3 (b) ec ec 4 (c) ec 3 (b) tan ec ( ) (e) tan sin (c) ( cot ec ) 8.3 PROPERTIES OF INVERSE TRIGONOMETRIC FUNCTIONS Property sin ( sinθ ) =θ, θ Solution : Let sinθ= Also ( ) θ= sin sin sin = = sin ( sin θ ) =θ Similarly, we can prove that (i) ( θ ) =θ, 0 θ (ii) tan ( tanθ ) =θ, <θ< Property (i) ec = sin (ii) cot = tan MATHEMATICS 47

MODULE - IV (iii) sec = Inverse Trigonometric Solution : (i) Let ec =θ = ec θ sin = θ θ= sin ec = sin (iii) sec =θ = sec θ = θ or θ= sec = (ii) Let cot =θ = cot θ tan = θ θ= tan cot = tan Property 3 (i) sin = sin (ii) tan =tan (iii) ( ) = Solution : (i) Let sin ( ) =θ = sinθ or =sinθ= sin( θ) θ= sin or θ=sin or sin =sin (ii) Let tan ( ) =θ = tan θ or =tanθ= tan( θ) θ=tan or tan =tan (iii) Let ( ) =θ = θ or =θ= ( θ) =θ = Property 4. (i) sin + = (ii) (iii) ec + sec = tan + cot = 48 MATHEMATICS

Inverse Trigonometric Soluton : (i) sin + = Let sin =θ = sinθ= θ or = θ (ii) Let θ+ = or sin + = cot =θ = cotθ= tan θ or tan cot + tan = (iii) Let ec =θ = ecθ= sec θ sec ec + sec = = θ or θ+ tan = = θ or θ+ sec = y Property 5 (i) + tan + tan y= tan y (ii) y tan tan y = tan + y Solution : (i) Let tan =θ, tan y =φ = tan θ, y = tan φ y We have to prove that + tan + tan y= tan y By substituting that above values on L.H.S. and R.H.S., we have The result holds. Simiarly (ii) can be proved. tanθ+ tan φ L.H.S. = θ+φ and R.H.S. = tan tan θtan φ = tan tan θ+φ =θ+φ= L.H.S. MODULE - IV MATHEMATICS 49

MODULE - IV Inverse Trigonometric Property 6 tan = sin = tan = + + (i) (ii) (iii) (iv) Let = tan θ Substituting in (i), (ii), (iii), and (iv) we get tan = tan ( tanθ ) = θ...(i) sin tan = sin θ tan + + tan θ tan θ = sin sec θ ( sin ) = sin θ θ = sin sinθ = θ...(ii) θ + + tan θ = tan sin θ θ = θ+ sin θ = θsin θ = ( θ ) = θ...(iii) tan = tan θ tan θ = tan tanθ = θ...(iv) From (i), (ii), (iii) and (iv), we get tan = sin tan = = + + Property 7 sin = = tan (i) = sec = cot = ec 50 MATHEMATICS

Inverse Trigonometric = sin = tan (ii) = ec = cot MODULE - IV Proof : Let sin (i) θ=, = sec =θ sinθ= tan θ=, secθ= sin =θ= = tan = sec, cot θ= and ecθ= = cot (ii) Let sinθ=, and ecθ= = ec =θ = θ = sin tan θ=, secθ=, cot θ= = tan = ec = sec MATHEMATICS 5

MODULE - IV Inverse Trigonometric Property 8 (i) sin + sin y = sin y + y (ii) + y= y y sin sin y = sin y y (iii) y = y + y (iv) Proof (i) : Let L.H.S. = θ+φ = sinθ, y = sin φ, then R.H.S. = sin ( sinθφ+ θsin φ) L.H.S. = R.H.S. sin = sin θ+φ =θ+φ (ii) Let = θ and y = φ (iii) Let L.H.S. = θ+φ R.H.S. = ( θφsinθsin φ) L.H.S. = R.H.S. = θ+φ =θ+φ = sinθ, y = sin φ L.H.S. = θφ R.H.S. = sin y y L.H.S. = R.H.S. = sin sinθ sin φsinφ sin θ [ ] = sin sinθφ θsin φ sin = sin θφ =θφ 5 (iv) Let = θ, y = φ L.H.S = θφ R.H.S. = [ θφ+ sinθsin φ] = θφ =θφ L.H.S. = R.H.S. MATHEMATICS

Inverse Trigonometric Eample 8.5 Evaluate : sin Soluton : Let 3 5 + sin 5 3 sin =θ and sin =φ, then 5 3 3 3 5 sinθ= and sinφ= 5 3 5 4 θ= and φ= 5 3 The given epression becomes [ θ+φ] Eample 8.6 Prove that = θφsinθsin φ 4 3 5 33 = = 5 3 5 3 65 MODULE - IV tan tan tan + = 7 3 9 Solution : Applying the formula : + y tan + tan y= tan y, we have + 7 3 0 tan + tan = tan = tan = tan 7 3 90 9 7 3 Eample 8.7 Prove that 4 33 + = 5 3 65 Applying the property + y= y y, we have 4 4 6 44 + = 5 3 5 3 5 69 33 = 65 MATHEMATICS 53

MODULE - IV Eample 8.8 Prove that tan = + Solution : Let = tan θ then Inverse Trigonometric tan θ = θ + tan θ L.H.S. = θ and R.H.S. = ( ) = θ=θ L.H.S. = R.H.S. Eample 8.9 Solve the equation Solution : Let tan = tan, > 0 + = tan θ, then tanθ tan = tan tan θ + tanθ tan tan θ = θ 4 θ= θ 4 θ= = 4 3 6 = tan = 6 3 54 Eample 8.0 Show that Solution : Let Substituting tan tan = + + 4 + + = θ, then θ= θ= = θ in L.H.S. of the given equation, we have tan + θ+ θ = + θ θ + + + MATHEMATICS

Inverse Trigonometric sin tan θ+ θ = θ sin θ θ+ sinθ = tan θsinθ + tan θ = tan tan θ MODULE - IV = tan tan +θ 4 = +θ 4 = + 4 CHECK YOUR PROGRESS 8.. Evaluate each of the following : (a) sin sin 3 (b) ( ) cot tan α+ cot α (c) y tan sin + + y + (d) tan tan 5 (e) tan tan 5 4. If + y =β, prove that yβ+ y = sin β 3. If + y+ z=, prove that + y + z + yz = 4. Prove each of the following : (a) (c) sin + sin = (b) 5 5 4 3 7 + tan = tan (d) 5 5 4 5 6 sin + sin + sin = 5 3 65 tan + tan + tan = 5 8 4 MATHEMATICS 55

MODULE - IV LET US SUM UP LET US SUM UP Inverse Trigonometric Inverse of a trigonometric function eists if we restrict the domain of it. (i) sin = y if siny = where, y (ii) = y if y = where,0 y (iii) tan = y if tan y = where R, < y< (iv) cot = y if cot y = where R,0< y< (v) sec = y if sec y = where, 0 y< or, < y (vi) ec = y if ec y = where,0 < y or, y < 0 Graphs of inverse trigonometric functions can be represented in the given intervals by interchanging the aes as in case of y = sin, etc. Properties : (i) sin ( sinθ ) =θ, tan ( tan ) θ =θ, ( tan tan θ ) =θ and ( ) (ii) ec = sin, cot = tan, sec = sin sin θ =θ (iii) sin = sin, tan = tan, = (iv) (v) (vi) sin + =, tan + cot =, ec + sec = + y tan + tan y= tan y, y tan tan y = tan + y tan = sin tan = = + + = = (vii) sin tan = sec cot ec = = (viii) sin ± sin y = sin y ± y 56 ± y = y y y (i) MATHEMATICS

Inverse Trigonometric SUPPORTIVE WEB SITES MODULE - IV http://www.wikipedia.org http://mathworld.wolfram.com TERMINAL EXERCISE. Prove each of the following : 3 8 77 (a) sin sin sin + = 5 7 85 (b) 3 tan tan + = 4 9 5 4 3 7 + tan = tan 5 5 (c). Prove each of the following : 3 tan + tan = tan (b) 5 (a) (c) tan tan tan + = 8 5 3 3. (a) Prove that sin = sin (b) Prove that ( ) = tan tan + = tan 3 + (c) Prove that = sin = 4. Prove the following : (a) tan = sin + sin (b) tan 4 = sin + 4 ab+ bc+ ca+ cot + cot + cot = 0 ab bc ca (c) 5. Solve each of the following : (a) tan tan 3 (c) + = (b) tan ( ) = tan ( ec) 4 + sin = 6 (d) cot cot ( + ) =, > 0 MATHEMATICS 57

MODULE - IV ANSWERS CHECK YOUR PROGRESS 8. Inverse Trigonometric. (a) 6 (b) 4 (c) 3 (d) 3 (e) 4. (a) 3 (b) 4 (c) (d) (e) 3. (a) + (b) 4 (c) 4 (d) 4 + (e) CHECK YOUR PROGRESS 8.. (a) (b) 0 (c) + y y TERMINAL EXERCISE (d) 5 (e) 7 7 5. (a) 6 (b) 4 (c) ± (d) 3 58 MATHEMATICS