# Simplify Inverse Trigonometric Functions

How to simplify expressions including inverse trigonometric functions for grade 12 maths. Questions with detailed solutions are also included.

## Question 1

Simplify the expressions:a) sin(arcsin(x)) and arcsin(sin(x))

b) cos(arccos(x)) and arccos(cos(x))

c) tan(arctan(x)) and arctan(tan(x))

__Solution__

a) sin and arcsin are inverse of each other and therefore the properties of inverse functions may be used to write

sin(arcsin(x)) = x , for -1 ≤ x ≤ 1

arcsin(sin(x)) = x , for x ∈ [-π/2 , π/2]

NOTE: If x in arcsin(sin(x)) is not in the interval [-π/2 , π/2], find θ in the interval [-π/2 , π/2] such that sin(x) = sin(θ) and then simplify arcsin(sin(x)) = θ

b) cos and arccos are inverse of each other and therefore the properties of inverse functions may be used to write

cos(arccos(x)) = x , for -1 ≤ x ≤ 1

arccos(cos(x)) = x , for for x ∈ [0 , π]

NOTE: If x in arccos(cos(x)) is not in the interval [0/2 , π], find θ in the interval [0 , π] such that cos(x) = cos(θ) and then simplify arccos(cos(x)) = θ

c) tan and arctan are inverse of each other and therefore the properties of inverse functions may be used to write

tan(arctan(x)) = x

arctan(tan(x)) = x , for x ∈ (-π/2 , π/2)

NOTE: If x in arctan(tan(x)) is not in the interval (-π/2 , π/2), find θ in the interval (-π/2 , π/2) such that tan(x) = tan(θ) and then simplify arctan(tan(x)) = θ

## Question 2

Express the following as algebraic expressions:sin(arccos(x)) and tan(arccos(x))

__Solution__

Let A = arccos(x). Hence

cos(A) = cos(arccos(x)) = x

Use right triangle with angle A such that cos(A) = x (or x / 1), find second leg and calculate sin(A) and tan(A)

.

sin(arccos(x)) = sin(A) = √(1 - x

^{2}) / 1 = √(1 - x

^{2}) for x ∈ [-1 , 1]

tan(arccos(x)) = tan(A) = √(1 - x

^{2}) / x for x ∈ [-1 , 0) ∪ (0 , 1]

## Question 3

Express the following as algebraic expressions:cos(arcsin(x)) and tan(arcsin(x))

__Solution__

Let A = arcsin(x). Hence

sin(A) = sin(arcsin(x)) = x

Use right triangle with angle A such that sin(A) = x (or x / 1), find second leg and calculate cos(A) and tan(A)

.

cos(arcsin(x)) = cos(A) = √(1 - x

^{2}) / 1 = √(1 - x

^{2}) for x ∈ [-1 , 1]

tan(arcsin(x)) = tan(A) = x / √(1 - x

^{2}) for x ∈ (-1 , 1)

## Question 4

Express the following as algebraic expressions:sin(arctan(x)) and cos(arctan(x))

__Solution__

Let A = arctan(x). Hence

tan(A) = tan(arctan(x)) = x

Use right triangle with angle A such that tan(A) = x (or x / 1), find hypotenuse and calculate sin(A) and cos(A)

.

sin(arctan(x)) = sin(A) = x / √(1 + x

^{2})

cos(arctan(x)) = cos(A) = 1 / √(1 + x

^{2})

## Question 5

Simplify the following expressions:a) arccos(0) , arcsin(-1) , arctan(-1)

b) sin(arcsin(-1/2)) , arccos(cos(π/2)) , arccos(cos(-π/2))

c) cos(arcsin(-1/2)) , arcsin(sin(π/3)) , arcsin(tan(3π/4))

d) arccos(tan(7π/4)) , arcsin(sin(13π/3)) , arctan(tan(-17π/4)) , arcsin(sin(9π/5))

__Solution__

a) Use definition.

arccos(0) = π/2 because cos(π/2) = 0 and π/2 is within range of arccos which is [0 , π]

arcsin(-1) = -π/2 because sin(-π/2) = -1 and -π/2 is within range of arcsin which is [-π/2 , π/2]

arctan(-1) = -π/4 because tan(-π/4) = -1 and -π/4 is within range of arctan which is (-π/2 , π/2)

b) Simplify inner functions then the outer functions using definitions.

sin(arcsin(-1/2)) = sin(-π/6) = -1/2

arccos(cos(π/2)) = arccos(0) = π/2

arccos(cos(-π/2)) = arccos(0) = π/2

c) Simplify inner functions then the outer functions using definitions.

cos(arcsin(-1/2)) = cos(-π/6) = √3/2

arcsin(sin(π/3)) = arcsin(√3/2) = π/3

arcsin(tan(3π/4)) = arcsin(-1) = -π/2

d) Simplify inner functions then the outer functions using definitions.

arccos(tan(7π/4)) = arccos(-1) = π

arcsin(sin(13π/3)) = arcsin(sin(4π + π/3)) = arcsin(sin(π/3)) = π/3

arctan(tan(- 17π/4)) = arctan(tan(- 4π-π/4)) = arctan(tan(- π/4)) = - π/4

arcsin(sin(9π/5)) = arcsin(sin(2π - π/5)) = arcsin(sin(- π/5)) = - π/5

## Question 6

Let A = arcsin(2/3) and B = arccos(-1/2). Find the exact value of sin(A + B).__Solution__

Use the indentity sin(A + B) = sin(A)cos(B) + cos(A)sin(B) to expand the given expression.

sin(A + B) = sin(arcsin(2/3))cos(arccos(-1/2)) + cos(arcsin(2/3))sin(arccos(-1/2))

Use the above indentities to simplify each term in the above expression.

sin(arcsin(2/3)) = 2/3 (we have used sin(arcsin(x)) = x))

cos(arccos(-1/2)) = -1/2 (we have used cos(arccos(x)) = x))

cos(arcsin(2/3)) = √(1 - (2/3)

^{2}) = √5/3 (we have used cos(arcsin(x)) = √(1 - x

^{2}))

sin(arccos(-1/2)) = √(1 - (- 1/2)

^{2}) = √3/2 (we have used sin(arccos(x)) = √(1 - x

^{2})) Substitute and calculate.

sin(A + B) = (2/3)(-1/2)+(√5/3)(√3/2) = -1/3 + √(15)/6

## Question 7

Write Y = sin(2 arcsin(x)) as an algebraic expression.__Solution__

Let A = arcsin(x). Hence Y may be written as

Y = sin (2 A)

Use the identity sin(2 A) = 2 sin(A) cos(A) to rewrite Y as folllows:

Y = 2 sin (A) cos(A) = 2 sin(arcsin(x)) cos(arcsin(x))

Use the identities sin(arcsin(x)) = x and cos(arcsin(x)) = √(1-x

^{2}) to rewrite Y as follows:

Y = 2 x √(1 - x

^{2})

## Question 8

Find the exact value of Y = sin(2 arctan(3/4)).__Solution__

Let A = arctan(3/4). Hence Y may be written as

Y = sin(2 A) = 2 sin(A) cos(A)

sin(A) = sin(arctan(3/4)) = (3/4) / √(1 + (3/4)

^{2}) = 3/5

cos(A) = cos(arctan(3/4)) = 1 / √(1 + (3/4)

^{2}) = 4 / 5

Y = 2 (3 / 5)(4 / 5) = 24 / 25

__More References and links__

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