Lecture 6 : Inverse Trigonometric Functions Inverse Sine
Inverse Sine Function (arcsin x = sin 1x) The trigonometric function sinxis not one-to-one functions, hence in order to create an inverse, we must restrict its domain The restricted sine function is given by f(x) = 8
Restricted Sine Function
Inverse Sine Function (arcsin x = sin 1x) We see from the graph of the restricted sine function (or from its derivative) that the function is one-to-one and hence has an inverse, shown in red in the diagram below Hp 2,1L H-p 4,-1 2 L H1,p 2L H-1 2,-p 4 L-p 2-p 4 p 4 p 2-1 5-1 0-0 5 0 5 1 0 1 5 This inverse function, f 1(x), is denoted by f 1
Inverse functions - University of Arkansas
arcsin(sinx) = xif 0
Section 55 Inverse Trigonometric Functions and Their Graphs
Section 5 5 Inverse Trigonometric Functions and Their Graphs DEFINITION: The inverse sine function, denoted by sin 1 x (or arcsinx), is de ned to be the inverse of the restricted sine function
Inverse trigonometric functions (Sect 76) Review
The derivative of arcsin is given by arcsin0(x) = 1 √ 1 − x2 Proof: For x ∈ [−1,1] holds arcsin0(x) = 1 sin0 arcsin(x) = 1 cos arcsin(x) For x ∈ [−1,1] we get arcsin(x) = y ∈ hπ 2, π 2 i, and the cosine is positive in that interval, then cos(y) = + q 1 − sin2(y), hence arcsin0(x) = 1 q 1 − sin2 arcsin(x) ⇒ arcsin 0(x) = 1
Inverse trigonometric functions (Sect 76)
The derivative of arcsin is given by arcsin0(x) = 1 √ 1−x2 Proof: For x ∈ [−1,1] holds arcsin0(x) = 1 sin0 arcsin(x) = 1 cos arcsin(x) For x ∈ [−1,1] we get arcsin(x) = y ∈ hπ 2, π 2 i, and the cosine is positive in that interval, then cos(y) = + q 1−sin2(y), hence arcsin0(x) = 1 q 1−sin2 arcsin(x) ⇒ arcsin 0(x) = 1 √ 1
The complex inverse trigonometric and hyperbolic functions
2 The inverse trigonometric functions: arcsin and arccos The arcsine function is the solution to the equation: z = sinw = eiw − e−iw 2i ∗In our conventions, the real inverse tangent function, Arctan x, is a continuous single-valued function that varies smoothly from − 1 2π to +2π as x varies from −∞ to +∞ In contrast, Arccotx
Formule trigonometrice a b a b c b a c - Math
53 arcsinx+arcsiny= 2 6 6 6 4 arcsin(x p 1 y2 + y 1 x2); daca xy 0 sau x2 + y2 1;
Lecture 23: Improper integrals - Harvard University
Solution: The antiderivative is arcsin(x) In this case, it is not the point x = 0 which produces the difficulty It is the point x = 1 Take a > 0 and evaluate Z 1−a 0 1 √ 1− x2 dx = arcsin(x)1−a 0 = arcsin(1− a)− arcsin(0) Now arcsin(1 − a) has no problem at limit a → 0 Since arcsin(1) = π/2 exists We get therefore the
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Inverse trigonometric functions (Sect. 7.6)
Today:Derivatives and integrals.
?Review: Definitions and properties. ?Derivatives. ?Integrals.Last class:Definitions and properties.
?Domains restrictions and inverse trigs. ?Evaluating inverse trigs at simple values. ?Few identities for inverse trigs.Review: Definitions and properties Remark:On certain domains the trigonometric functions are invertible.1 yy = sin(x) x- π / 2π / 2 -1 1 y xy = cos(x)π0π / 2
-1 p / 2x y = tan(x)y - p / 2 0xy y = csc(x) - π / 2π / 2 -11 y x1 -10π / 2π
y = sec(x) p / 2 y x0y = cot(x)
pReview: Definitions and properties
Remark:The graph of the inverse function is a reflection of the original function graph about they=xaxis.y = arcsin(x) xπ / 2 - π / 2 1-1y y = arccos(x)0π / 2π
y x -11 y x - π / 2 π / 2 y = arctan(x)y = arccsc(x)y -10 1π / 2
- π / 2 xy = arcsec(x) -1 10π / 2π y x y0π / 2π
x y = arccot(x)Review: Definitions and propertiesTheorem
For all x?[-1,1]the following identities hold,arccos(x) + arccos(-x) =π,arccos(x) + arcsin(x) =π2
.Proof:arccos(-x) 1y (θ)x = cos(π-θ)-x = cosπ - θ x arccos(x)arccos(x) 1y (θ)x = cosxπ/2 - θ(π/2-θ)x = sinarcsin(x)
Review: Definitions and properties
Theorem
For all x?[-1,1]the following identities hold,arcsin(-x) =-arcsin(x), arctan(-x) =-arctan(x), arccsc(-x) =-arccsc(x).Proof:y = arcsin(x) xπ / 2 - π / 2 1-1y y x - π / 2 π / 2 y = arctan(x)y = arccsc(x)y -10 1π / 2
- π / 2 xInverse trigonometric functions (Sect. 7.6)Today:Derivatives and integrals.
?Review: Definitions and properties. ?Derivatives. ?Integrals.Derivatives of inverse trigonometric functions
Remark:Derivatives inverse functions can be computed with f-1??(x) =1f ??f-1(x)?.TheoremThe derivative ofarcsinis given byarcsin
?(x) =1⎷1-x2.Proof:Forx?[-1,1] holds
arcsin ?(x) =1sin ??arcsin(x)?= 1cos ?arcsin(x)?Forx?[-1,1] we get arcsin(x) =y??π2 ,π2 ,and the cosine is positive in that interval,then cos(y) = +?1-sin2(y),hence arcsin ?(x) =1?1-sin2?arcsin(x)??arcsin
?(x) =1⎷1-x2.Derivatives of inverse trigonometric functionsTheorem
The derivative of inverse trigonometric functions are: arcsin ?(x) =1⎷1-x2,arccos?(x) =-1⎷1-x2,|x|?1, arctan ?(x) =11 +x2,arccot?(x) =-11 +x2,x?R, arcsec ?(x) =1|x|⎷x2-1,arccsc?(x) =-1|x|⎷x
2-1,|x|?1.Proof:arctan
?(x) =1tan ??arctan(x)?,tan ?(y) =cos2(y) + sin2(y)cos2(y)tan
?(y) = 1 + tan2(y),y= arctan(x),?arctan ?(x) =11 +x2.Derivatives of inverse trigonometric functions
Proof:arcsec
?(x) =1sec ??arcsec(x)?,for|x|?1.Theny= arcsec(x) satisfiesy?[0,π]- {π/2}.Recall, sec ?(y) =?1cos(y)? sin(y)cos2(y),sin(y) = +?1-cos2(y),sec
?(y) =?1-cos2(y)cos 2(y)=1|cos(y)|?1-cos2(y)|cos(y)|,sec
?(y) =1|cos(y)|?1 cos2(y)-1=|sec(y)|?sec
2(y)-1.We conclude:arcsec
?(x) =1|x|⎷x2-1.Derivatives of inverse trigonometric functions
Example
Compute the derivative ofy(x) = arcsec(3x+ 7).Solution:Recall the main formula: arcsec ?(u) =1|u|⎷u2-1.Then, chain rule implies,y
?(x) =3|3x+ 7|?(3x+ 7)2-1.?Example
Compute the derivative ofy(x) = arctan(4ln(x)).Solution:Recall the main formula: arctan ?(u) =11 +u2.Therefore, chain rule implies, y ?(x) =1?1 +?4ln(x)?2?4x?y
?=4x ?1 + 16ln2(x)?.?Inverse trigonometric functions (Sect. 7.6)
Today:Derivatives and integrals.
?Review: Definitions and properties. ?Derivatives. ?Integrals.Integrals of inverse trigonometric functions Remark:The formulas for the derivatives of inverse trigonometric functions imply the integration formulas.TheoremFor any constant a?= 0holds,?
dx⎷a2-x2= arcsin?xa
+c,|x|2-a2=1a
arcsec? ???xa +c,|x|>a>0.Proof:(For arcsine only.)y(x) = arcsin?xa +c,then y ?(x)= 1? 1-x2a 21a=|a|⎷a