Simplify Expressions Including Inverse Trigonometric Functions



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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)

    triangle 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)

    triangle 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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