# Simplify Expressions Including Inverse Trigonometric Functions

| How to simplify expressions including inverse trigonometric functions for grade 12 maths. Questions with detailed solutions are also included.
## Question 1Simplify 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 2Express 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 3Express 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 4Express 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 5Simplify 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 6Let 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 7Write 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 8Find 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 Inverse Trigonometric Functions |