Inverse trigonometric functions (Sect 76) Review

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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 -10

π / 2π

y = sec(x) p / 2 y x

0y = cot(x)

Review: 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 -1

0 1π / 2

- π / 2 xy = arcsec(x) -1 10π / 2π y x y

0π / 2π

x y = arccot(x)Review: Definitions and properties

Theorem

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 = cosx

π/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 -1

0 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)?.Theorem

The 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 functions

Theorem

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|⎷x

2-1,arccsc?(x) =-1|x|⎷x

2-1,|x|?1.Proof:arctan

?(x) =1tan ??arctan(x)?,tan ?(y) =cos2(y) + sin2(y)cos

2(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)cos

2(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 cos

2(y)-1=|sec(y)|?sec

2(y)-1.We conclude:arcsec

?(x) =1|x|⎷x

2-1.Derivatives of inverse trigonometric functions

Example

Compute the derivative ofy(x) = arcsec(3x+ 7).Solution:Recall the main formula: arcsec ?(u) =1|u|⎷u

2-1.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)?.?