Simplify Expressions Including Inverse Trigonometric Functions

How to simplify expressions including inverse trigonometric functions for grade 12 math. Detailed solutions are also included.

  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


  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)

    trangle for question 2.



    sin(arccos(x)) = sin(A) = √(1 - x2) / 1 = √(1 - x2)     for x ∈ [-1 , 1]

    tan(arccos(x)) = tan(A) = √(1 - x2) / x     for x ∈ [-1 , 0) ∪ (0 , 1]

  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)

    trangle for question 3.



    cos(arcsin(x)) = cos(A) = √(1 - x2) / 1 = √(1 - x2)     for x ∈ [-1 , 1]

    tan(arcsin(x)) = tan(A) = x / √(1 - x2)     for x ∈ (-1 , 1)

  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)

    trangle for question 4.



    sin(arctan(x)) = sin(A) = x / √(1 + x2)

    cos(arctan(x)) = cos(A) = 1 / √(1 + x2)

  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

  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 - x2))

    sin(arccos(-1/2)) = √(1 - (- 1/2)2) = √3/2          (we have used sin(arccos(x)) = √(1 - x2)) Substitute and calculate.

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

  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-x2) to rewrite Y as follows:

    Y = 2 x √(1 - x2)

  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


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