Lecture 6 : Inverse Trigonometric Functions Inverse Sine

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Lecture 6 : Inverse Trigonometric Functions

Inverse Sine Function (arcsin x =sin1x)The trigonometric function sinxis not one-to-one functions, hence in order to create an inverse, we must restrict its domain.

The restricted sine functionis given by

f(x) =8 :sinx2 x2 undened otherwise

We have Domain(f) = [2

;2 ] and Range(f) = [1;1].Hp6,12LH5p6,12L-p-p 2 p 2 p-1.0-0.50.51.0y=sinx-p 2-p 4 p 4 p 2

-1.0-0.50.51.0y=fHxLWe 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.Hp2,1L H -p 4 -1 2 L

H1,p2L

H -1 2 -p 4 L-p 2-p 4 p 4 p 2 -1.5-1.0-0.50.51.01.5This inverse function,f1(x), is denoted byf

1(x) = sin1xor arcsinx:Properties ofsin1x.

Domain(sin

1) = [1;1] and Range(sin1) = [2

Sincef1(x) =yif and only iff(y) =x;we have:

1 sin

1x=yif and only if sin(y) =xand2

y2 :Sincef(f1)(x) =x f1(f(x)) =xwe have:sin(sin

1(x)) =xforx2[1;1] sin1(sin(x)) =xforx22

;2 :from the graph: sin

1xis an odd function and sin1(x) =sin1x:

ExampleEvaluate sin1

1p2 using the graph above.

ExampleEvaluate sin1(p3=2), sin1(p3=2),

ExampleEvaluate sin1(sin).

ExampleEvaluate cos(sin1(p3=2)).

ExampleGive a formula in terms ofxfor tan(sin1(x))

Derivative ofsin1x.d

sin1x=1p1x2;1x1:ProofWe have sin1x=yif and only if siny=x. Using implicit dierentiation, we get cosydydx

= 1 ordydx =1cosy:

Now we know that cos

2y+ sin2y= 1, hence we have that cos2y+x2= 1 and

cosy=p1x2 2 and ddx sin1x=1p1x2:If we use the chain rule in conjunction with the above derivative, we get d dx sin1(k(x)) =k0(x)p1(k(x))2; x2Dom(k) and1k(x)1:ExampleFind the derivativeddx sin1pcosx Inverse Cosine FunctionWe can dene the function cos1x= arccos(x) similarly. The details are given at the end of this lecture.

Domain(cos

1) = [1;1] and Range(cos1) = [0;].cos

1x=yif and only if cos(y) =xand 0y:cos(cos

1(x)) =xforx2[1;1] cos1(cos(x)) =xforx20;:It is shown at the end of the lecture that

ddx cos1x=ddx sin1x=1p1x2 and one can use this to prove thatsin

1x+ cos1x=2:

Inverse Tangent Function

The tangent function is not a one to one function, however we can also restrict the domain to construct

a one to one function in this case.

The restricted tangent functionis given by

h(x) =8 :tanx2 < x <2 undened otherwise

We see from the graph of the restricted tangent function (or from its derivative) that the function is

one-to-one and hence has an inverse, which we denote byh

1(x) = tan1xor arctanx:3

Hp4,1L-p

2-p 4 p 4 p 2 2-p 4 p 4 p 2 y=arctanHxLProperties oftan1x.

Domain(tan

1) = (1;1) and Range(tan1) = (2

Sinceh1(x) =yif and only ifh(y) =x;we have:tan

1x=yif and only if tan(y) =xand2

< y <2 :Sinceh(h1(x)) =xandh1(h(x)) =x;we have:tan(tan

1(x)) =xforx2(1;1) tan1(tan(x)) =xforx2

2 ;2 :Frpm the graph, we have:tan

1(x) =tan1(x):Also, since lim

x!(2 )tanx=1and lim x!(2 +)tanx=1; we havelim x!1tan1x=2andlim x!1tan1x=2

ExampleFind tan1(1) and tan1(1p3

ExampleFind cos(tan1(1p3

Derivative oftan1x.d

dx tan1x=1x

2+ 1;1< x <1:4

ProofWe have tan1x=yif and only if tany=x. Using implicit dierentiation, we get sec2ydydx = 1 ordydx =1sec

2y= cos2y:

Now we know that cos

2y= cos2(tan1x) =11+x2:proving the result.If we use the chain rule in conjunction with the above derivative, we get

d dx tan1(k(x)) =k0(x)1 + (k(x))2; x2Dom(k)ExampleFind the domain and derivative of tan1(lnx)

Domain = (0;1)

ddx tan1(lnx) =1x

1 + (lnx)2=1x(1 + (lnx)2)

Integration formulas

Reversing the derivative formulas above, we getZ

1p1x2dx= sin1x+C;Z1x

2+ 1dx= tan1x+C;Example

Z1p9x2dx=

Z 13 q1x29 dx=Z13 q1x29 dx=13 Z 1q 1x29 dx

Letu=x3

, thendx= 3duand

Z1p9x2dx=13

3p1u2du= sin1u+C= sin1x3

Example

Z 1=2

011 + 4x2dx

Letu= 2x, thendu= 2dx,u(0) = 0,u(1=2) = 1 and

Z 1=2

011 + 4x2dx=12

Z 1

011 +u2dx=12

tan1uj10=12 [tan1(1)tan1(0)] 12 [4 0] =8 5

The restricted cosine functionis given by

g(x) =8 :cosx0x undened otherwise

We have Domain(g) = [0;] and Range(g) = [1;1].54321-1-2-3!2-3!-5!2-2!-3!2-!-!2!2!3!22!5!2We see from the graph of the restricted cosine function (or from its derivative) that the function is

one-to-one and hence has an inverse,g

1(x) = cos1xor arccosx3!2!!2-!2-!-4-224fx() = cos-1x()6

Domain(cos

1) = [1;1] and Range(cos1) = [0;].

Recall from the denition of inverse functions:

1(x) =yif and only ifg(y) =x:cos

1x=yif and only if cos(y) =xand 0y:g(g1(x)) =x g1(g(x)) =xcos(cos

1(x)) =xforx2[1;1] cos1(cos(x)) =xforx20;:Note from the graph thatcos

1(x) =cos1(x).

cos

1(p3=2) =and cos

1(p3=2) =You can use either chart below to nd the correct angle between 0 and.:tan(cos

1(p3=2)) =tan(cos

1(x)) =Must draw a triangle with correct proportions:1xcos θ = xθtan(cos-1x) = tan θ = 1-x2xcos-1x = θ1-x21xcos θ = xθ7

d dx

cos1x=1p1x2;1x1:ProofWe have cos1x=yif and only if cosy=x. Using implicit dierentiation, we getsinydydx

= 1 ordydx =1siny:

Now we know that cos

2y+ sin2y= 1, hence we have that sin2y+x2= 1 and

siny=p1x2 and ddx cos1x=1p1x2:

Notethatddx

cos1x=ddx sin1x. In fact we can use this to prove thatsin

1x+ cos1x=2.

If we use the chain rule in conjunction with the above derivative, we get d dx cos1(k(x)) =k0(x)p1(k(x))2; x2Dom(k) and1k(x)1:8quotesdbs_dbs16.pdfusesText_22

[PDF] Lecture 6 : Inverse Trigonometric Functions Inverse Sine

Section 55 Inverse Trigonometric Functions and Their Graphs