MTH 112 Elementary Functions Chapter 6 Trigonometric Identities, Inverse Functions, and Equations

MTH 112Elementary Functions

Chapter 6

Trigonometric Identities, Inverse Functions, and Equations

Section 4Inverses of the Trigonometric Functions

Inverse of a Function?

Function: f = { (a,b) | aD bR f(a) = b }• D = Domain of the Function• R = Range of the Function

Inverse: f -1 = { (b,a) | (a,b)f and f -1(b) = a}Basic relationship: f -1(f(x)) = x and f(f -1(x)) = x Does the inverse of a function exist?

• Only if the function is one-to-one.• That is, if f(a) = f(b) then a = b for all a & b in the domain of f.

Examples

Does the inverse of y = x2 exist?– No, because a2 = b2 does not imply that a = b.

Does the inverse of y = x2 where x 0 exist?– Yes!

Do these two functions have the same range?– Yes!

Note that if the inverse of a function does not exist, sometimes the inverse of the original function with a restricted domain and same range does exist.

The Graph of theInverse of a Function

What is the relationship between the graph of a function and the graph of its inverse?– Reflection about the line y = x.

Why?– Each ordered pair (a,b) of the

function corresponds to an ordered pair (b,a) of the inverse.

y = x

(a,b)

(b,a)

bb

a

a

Inverses of Trig Functions

Do the inverses of the trig functions exist?– NO!

Try the following on a calculator.sin-1(sin(1)) sin-1(sin(2))sin-1(sin(4))sin-1(sin(6))sin-1(sin(-1))sin-1(sin(-2))sin-1(sin(-4))sin-1(sin(-6))

Note: 1 radian 57° 2 radians 115° 4 radians 229° 6 radians 344°



Try the following on a calculator.sin-1(sin(1)) = 1sin-1(sin(2)) 1.14sin-1(sin(4)) -0.86sin-1(sin(6)) -0.28sin-1(sin(-1)) = -1sin-1(sin(-2)) -1.14sin-1(sin(-4)) -0.86sin-1(sin(-6)) -0.28




Try the following on a calculator.cos-1(cos(1)) cos-1(cos(2))cos-1(cos(4))cos-1(cos(6))cos-1(cos(-1))cos-1(cos(-2))cos-1(cos(-4))cos-1(cos(-6))




Try the following on a calculator.cos-1(cos(1)) = 1cos-1(cos(2)) = 2cos-1(cos(4)) 2.28cos-1(cos(6)) 0.28cos-1(cos(-1)) = 1cos-1(cos(-2)) = 2cos-1(cos(-4)) 2.28cos-1(cos(-6)) 0.28



Try the following on a calculator.sin(sin-1(1))sin(sin-1(2))sin(sin-1(-1))sin(sin-1(-2))sin(sin-1(0.5))sin(sin-1(-0.5))


Try the following on a calculator.sin(sin-1(1)) = 1sin(sin-1(2)) = errorsin(sin-1(-1)) = -1sin(sin-1(-2)) = errorsin(sin-1(0.5)) = 0.5sin(sin-1(-0.5)) = -0.5

Try the following on a calculator.cos(cos-1(1))cos(cos-1(2))cos(cos-1(-1))cos(cos-1(-2))cos(cos-1(0.5))cos(cos-1(-0.5))


Try the following on a calculator.sin(sin-1(1)) = 1sin(sin-1(2)) = errorsin(sin-1(-1)) = -1sin(sin-1(-2)) = errorsin(sin-1(0.5)) = 0.5sin(sin-1(-0.5)) = -0.5

Try the following on a calculator.cos(cos-1(1)) = 1cos(cos-1(2)) = errorcos(cos-1(-1)) = -1cos(cos-1(-2)) = errorcos(cos-1(0.5)) = 0.5cos(cos-1(-0.5)) = -0.5

-1

-0.8

-0.6

-0.4

-0.20

0.2

0.4

0.6

0.8

1

-6 -4 -2 2 4 6x- 2-2

(/2, 1)

(3/2, -1)(-/2, -1)

(-3/2, 1)y = sin x

y = sin-1x = arcsin x

Begin with y = sin x.

Restrict the domain to [-/2, /2].

This gives a 1-1 function with the same range as the original function.

y = sin-1x = arcsin x

Therefore …• y = sin-1x x = sin y, where y [-/2, /2]• Domain: [-1, 1]• Range: [-/2, /2]

-1.5

-1

-0.5

0

0.5

1

1.5

-1 -0.8 -0.6 -0.4 -0.2 0.2 0.4 0.6 0.8 1x

y = cos-1x = arccos x

-1

-0.8

-0.6

-0.4

-0.20

0.2

0.4

0.6

0.8

1

-6 -4 -2 2 4 6x/2

-/2

3/2-3/2

(2, 1)

(, -1)(-, -1)

(-2, 1) (0, 1)

y = cos x

Begin with y = cos x.

Restrict the domain to [0, ].


y = cos-1x = arccos x

Therefore …• y = cos-1x x = cos y, where y [0, ]• Domain: [-1, 1]• Range: [0, ]

0

0.5

1

1.5

2

2.5

3

-1 -0.8 -0.6 -0.4 -0.2 0.2 0.4 0.6 0.8 1x

y = tan-1x = arctan x

Begin with y = tan x.

Restrict the domain to (-/2, /2).


-2

-1

0

1

2

-6 -4 -2 2 4 6/2-/2 3/2-

3/2

(-, 0) (, 0)(0, 0)(-2, 0) (2, 0)

y = tan x

y = tan-1x = arctan x

Therefore …• y = tan-1x x = tan y, where y (-/2, /2)• Domain: (-, )• Range: (-/2, /2)

-1.5

-1

-0.5

0

0.5

1

1.5

-4 -2 2 4x

Likewise, the other three …

y = sec-1x = arcsec x x = sec y– Domain: (-, -1] [1, )– Range: [0, /2) (/2, ]

y = csc-1x = arccsc x x = csc y– Domain: (-, -1] [1, )– Range: [-/2, 0) (0, /2]

y = cot-1x = arccot x x = cot y– Domain: (-, )– Range: [-/2, 0) (0, /2]

• Why not (0, )? Because of calculators!

sec-1x, csc-1x & cot-1xOn a Calculator

y = sec-1x x = sec y x = 1/cos y cos y = 1/x y = cos-1(1/x)

sec-1x, csc-1x & cot-1xOn a Calculator

Therefore …

• y = sec-1x y = cos-1(1/x)

• y = csc-1x y = sin-1(1/x)

• y = cot-1x y = tan-1(1/x)

Composition of Trig Functionswith Inverse Trig Functions

Under what conditions does sin-1(sin x) = x ?– The range of Inverse Sine function is [-/2, /2]– Therefore, x [-/2, /2].


sin-1(sin x) = x x [-/2, /2]cos-1(cos x) = x x [0, ]tan-1(tan x) = x x (-/2, /2)

What about something like .. sin-1(sin 5/6) = ? (5/6 is not in the above interval)

= sin-1(1/2) = /6

• It can still be evaluated, it’s just not equal to 5/6.


Under what conditions does sin(sin-1x) = x ?– The domain of Inverse Sine function is [-1, 1]– Therefore, x [-1, 1].


sin(sin-1x) = x x [-1, 1]cos(cos-1x) = x x [-1, 1]tan(tan-1x) = x x (-, )

What about something like .. sin(sin-1(5/3)) = ? (5/3 is not in the above interval)

• It can not be evaluated, because 5/3 is not in the domain of the inverse sine function.


What about something like ... sin-1(cos x)– If cos x is a known value, evaluate it and then find

the angle whose sine is this value.– If cos x is not a known value, use a calculator.

Example: sin-1(cos(4/3)) = sin-1(-1/2) = -/6


What about something like ... sin(cos-1x)– This can always be evaluated without a calculator.– Remember: cos-1x represents an acute angle (if x > 0)!

• Draw a triangle where cos-1x is one of the acute angles.• Using the definition, if the adjacent side is x, then the

hypotenuse will be 1.• The opposite side will then be

cos-1xx

1 21 xTherefore,

21 1cossin xx

21 x


Another variation ... same approach: csc(tan-1x)– tan-1x represents an acute angle (if x > 0)!

• Draw a triangle where tan-1x is one of the acute angles.• Using the definition, if the opposite side is x, then the

adjacent side will be 1.• The hypotenuse will then be

tan-1xx

1

21 x Therefore,

x

xx

21 1

tancsc

21 x


21 1)sin(cos xx 21 1)cos(sin xx

1)tan(sec 21 xx 21 1)sec(tan xx

21 1)csc(cot xx 1)cot(csc 21 xx

Other combinations?

MTH 112 Elementary Functions Chapter 6 Trigonometric Identities, Inverse Functions, and Equations

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