Inverse Trig Functions

Transcription

1 Inverse Trig Functions Previously in Math 30 DEFINITION 24 A function g is the inverse of the function f if g( f ()) = for all in the omain of f 2 f (g()) = for all in the omain of g In this situation g is enote by f an is calle f inverse" The Key Fact on the Eistence of Inverses THEOREM 24 f has an inverse f is one-to-one f passes the HLT The Graph of f Now suppose that y = f () has an inverse, f () an assume that a is in the omain of f an that f (a) = b Then using the efinition of inverse: f (a) = b f ( f (a)) = f (b) a Inverse = f (b) In other wors f (a) = b f (b) = a or (a, b) on the graph of f (b, a) is on the graph of f In other wors, f an f have their an y coorinates switche An because the an y coorinates are switche Domain of f = Range of f Range of f = Domain of f If f is one-to-one, we can obtain the graph of f by interchanging the an y coorinates If we raw the iagonal line y = an use it as a mirror, notice that the an y aes are reflecte into each other across the line This is just another way of saying that the an y coorinates have been switche So to obtain the graph of f all we nee to o is to reflect the graph of f in the iagonal line y =, as shown to the right Introuction f f None of the trig functions have inverses because none of them pass the horizontal line test Their values repeat every 2π units or every π units (tangent, cotangent) pi π/2 π/2 π 3π/2 2π

2 2 The Inverse Sine Function However, if we restrict the omain of the sine function (or any of the other trig functions) we can make the function one-to-one on the restricte interval The figure on the left below shows sin restricte to the interval [ π/2, π/2] where it is, inee, one-to-one (passes HLT) So it has an inverse there, which we have graphe in re the figure on the right π/2 π/2 π/2 π/2 The inverse sine function is enote by arcsin Your tet uses sin, but most stuents fin arcsin less confusing, an that s what we will generally use in this course Since the omain an range of the sine an inverse sine functions are interchange, we have the omain of arcsin is the range of the restricte sin : [, ] the range of arcsin is the omain of the restricte sin : [ π/2, π/2] This is very important It says that the output of the inverse sine function is a number (an angle) between π/2 an π/2 Notice since the arcsine function unoes the sine function, we get some familiar values: arcsin( ) = π/2 since sin( π/2) = Or arcsin(/2) = π/6 since sin(π/6) = /2 Or arcsin( 3/2) = π/3 since sin(π/3) = 3/2 EXAMPLE 24 Normally when we calculate f ( f ()) we get because the two functions uno each other The same is true here, if the omain of sin is appropriately restricte to [ π/2, π/2] For eample, arcsin(sin(π/4)) = arcsin( 2/2) = π/4 But if we take a value outsie of the restricte omain [ π/2, π/2] of the sine function arcsin(sin(3π/4)) = arcsin( 2/2) = π/4 Or arcsin(sin(3π)) = arcsin(0) = 0 The two functions o not uno each other since the arcsine function can only return values (or angles) between π/2 an π/2 The Inverse Cosine Function We can restrict the omains of the other trig functions so that they, too, have inverses The figure on the left below shows cos restricte to the interval [0, π] where it is, inee, one-to-one So it has an inverse there, which we have graphe

3 math 30, calculus i 3 in re the figure on the right π π π π The inverse cosine function is enote by arccos Since the omain an range of the cosine an inverse cosine functions are interchange, we have the omain of arccos is the range of the restricte cos : [, ] the range of arccos is the omain of the restricte cos : [0, π] EXAMPLE 242 Again we have to be careful about calculating the composites of these inverse functions They are only inverses when the inputs are in the correct omains For eample, arccos(cos(π/4)) = arccos( 2/2) = π/4 But if we take a value outsie of the restricte omain [0, π] of the cosine function Or arccos(cos( π/4)) = arccos( 2/2) = π/4 arccos(cos(3π)) = arccos( ) = π The two functions o not always uno each other since the inverse cosine function can only return values between 0 an π The Inverse Tangent Function The figure on the left below shows tan restricte to the interval ( π/2, π/2) where it is, inee, one-to-one So it has an inverse there, which we have graphe in re the figure on the right π/2 π/2 π/2 π/2 π/2 π/2 The inverse tangent function is enote by arctan Since the omain an range of the tangent an inverse tangent functions are interchange, we have the omain of arctan is the range of the restricte tan : (, )

4 4 the range of arctan is the omain of the restricte tan : ( π/2, π/2) EXAMPLE 243 Again we have to be careful about calculating the composites of these inverse functions They are only inverses when the inputs are in the correct omains For eample, arctan(tan(π/4)) = arctan() = π/4 But if we take a value outsie of the restricte omain ( π/2, π/2) of the tangent function arctan(tan(3π/4)) = arctan( ) = π/4 Or arctan(tan(3π)) = arctan(0) = 0 The two functions o not always uno each other since the inverse tangent function can only return values between π/2 an π/2 We will concentrate only on the the three inverse functions iscusse above I will leave it to you to rea about the other inverse trig functions in your tet Evaluation Using Triangles Drawing appropriate right triangles can help evaluate complicate epressions involving the inverse trig functions EXAMPLE 244 Evaluate cos(arcsin ) SOLUTION Remember that arcsin = θ where θ is just the angle whose sine is We want the cosine of this same angle So let s raw a right triangle with angle θ whose sine is Since the sine function is opp hyp we can use the triangle below θ 2 + y 2 = y = 2 y Notice sin θ = = So arcsin = θ (θ is the angle whose sine is ) So cos(arcsin ) = cos(θ) = y = 2 = 2 EXAMPLE 245 Evaluate sec(arctan ) SOLUTION This time we raw a triangle whose tangent is z So θ z 2 = z = + 2 sec(arctan ) = sec(θ) = z = + 2 EXAMPLE 246 Evaluate sin(arccos 2/5) SOLUTION This time we raw a triangle whose cosine is 2/5 5 So θ = 5 2 = = 2 sin(arccos 2/5) = sin(θ) = 5 = 2 5 YOU TRY IT 24 Evaluate sin(arctan ) an cos(arcsin 3/4))

5 math 30, calculus i 5 Derivatives of arcsin an arctan Surprisingly, it is relatively easy to etermine the erivatives of the inverse trig functions, assuming that they are ifferentiable We will use implicit ifferentiation (really just the chain rule in isguise) just as we i when we figure out the erivative of ln Let s first etermine the erivative of y = arcsin for π 2 π 2 We want to fin y First apply the inverse: y = arcsin sin(y) = sin(arcsin ) = Now take the erivative using implicit ifferentiation on the left: D [sin(y)] = D [] cos(y) y = Solve for y y = cos(y) = cos(arcsin ) But in Eample 244 we foun that cos(arcsin ) = 2 so we have That is y = cos(arcsin ) = 2 (arcsin ) = 2 The erivative of y = arctan for π 2 < < π 2 is etermine in a similar fashion We want to fin y First apply the inverse: y = arctan tan(y) = tan(arctan ) = Now take the erivative using implicit ifferentiation on the left: D [tan(y)] = D [] sec 2 (y) y = Solve for y y = sec 2 (y) = sec 2 (arctan ) But in Eample 245 we foun that sec(arctan ) = + 2 so we have sec 2 (arctan ) = + 2 Therefore y = sec 2 (arctan ) = + 2 That is (arctan ) = + 2 YOU TRY IT 242 (Etra Creit) Determine the formula for the erivative of arccos using the metho above Show your work Keep going an fin the erivatives of the remaining three inverse trig functions Again show your work

6 6 Chain Rule Versions The chain rule versions of both erivative formulas are: (arcsin u) = u u 2 u (arctan u) = + u 2 EXAMPLE 247 Let s use these formulas to fin the erivatives of the following: (arctan e3 ) = + (e 3 ) 2 3e3 = 3e3 + e 6 (u = e3 ) (arcsin 32 ) = (3 2 ) 2 6 = (u = 32 ) (earctan 3 ) = e arctan = 3earctan (sin 2 arctan 52 ) = 2 cos 2 arctan sin 2 = 2 cos 2 arctan sin (ln arcsin 3 ) = arcsin 3 3 = (arcsin 3) 9 2 D ( arcsin(ln 3)) = [ln(3)] = [ln(3)] 2 (u = ln(3)) YOU TRY IT 243 Fin the erivatives of these functions: (a) arctan(62 )] (b) [arcsin( )] (e) [arctan(ln 6 )] (f ) [arcsin(6esin )] (i) (ln arctan + e4 ) The answers are on the net page (c) [arctan(e2 ] (g) (e2 arcsin 2 ) () [arcsin(arcsin )] (h) [(arcsin 2)(tan 52 )]

7 math 30, calculus i 7 Answers Answers to you try it 243 (a) 2 (arctan(62 )) = 2 = (b) (arcsin( )) = 2 /2 = 2 (c) () (e) (f ) (g) (h) (i) (arctan(e2 )) = 2e2 + e 4 (arcsin(arcsin )) = (arcsin ) 2 2 [arctan(ln 6 )] = + (ln 6 ) = [ + (ln 6 ) 2 ] (arcsin(6esin )) = (6e sin ) 2 (6esin )(cos ) = (e2 arcsin 2 ) = (e 2 arcsin 2 4e2 arcsin 2 ) 2 2 = ( 2 ) 2 4 [arcsin 2(tan 52 )] = (ln arctan e4 + ) = 2 tan (arcsin 2)0 sec2 (5 2 ) arctan e cos esin (6e sin ) 2 + (e 4 + ) 2 e = 4 3 e 4 + (arctan e 4 + )( + e )