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Mathematics Dictionary
Dr. K. G. Shih
Invesre Trigonometric Functions
Subjects


  • TR 08 00 | - y = arcsin(x)
  • TR 08 01 | - y = arcsin(x)
  • TR 08 02 | - y = arccos(x)
  • TR 08 03 | - y = arctan(x)
  • TR 08 04 | - arcsin(x) + arcsin(y) = arcsin(x*Sqr(1-y^2) + y*Sqr(1-x^2))
  • TR 08 05 | - arccos(x) + arccos(y) = arccos(x*y + Sqr(1-x^2)*Sqr(1-y^2))
  • TR 08 06 | - arctan(x) + arctan(y) = arctan((x + y)/(1 - x*y))
  • TR 08 07 | - Express arccos(x) in terms of atn(y)
  • TR 08 08 | - arcsin(sin(A)+sin(B)) + arcsin(sin(A)-sin(B)) = pi/2, sin(A)^2+sin(B)^2 = ?
  • TR 08 09 | - arcsin(4/5) + arcsin(5/13) + arcsin(16/65) = pi/2
  • TR 08 10 | - Prove that 2*arctan(1/3) + arctan(1/7) = 45 degrees
  • TR 08 11 | - More arcsin(x) and arctan(x)
    • Answers


    TR 08 01. x = arcsin(y)

    Defintion
    • If y = sin(x) then the inverse of sine function is x = arcsin(y).
    Composite function properteis
    • sin(arcsin(x)) = x
    • arcsin(sin(A)) = A.
    Example : Solve arcsin(-1/2) = x.
    • By defintion sin(x) = -1/2.
    • Hence x = pi + pi/6. or x = 2*pi - pi/6.
    • Gerenal solution : x = 2*n*pi - pi/6 or x = (2*n + 1)*pi + pi/6.

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    TR 08 02. x = arccos(y)

    Defintion
    • If y = cos(x) then the inverse of cosine function is x = arccos(y).
    Composite function properteis
    • cos(arccos(x)) = x
    • arccos(cos(A)) = A.
    Example : Solve arccos(1/2) = x.
    • By defintion cos(x) = 1/2.
    • Hence x = pi/3. or x = 2*pi - pi/3.
    • Gerenal solution : x = 2*n*pi + pi/3 or x = 2*n*pi - pi/3.

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    TR 08 03. x = arctan(y)

    Defintion
    • If y = tan(x) then the inverse of thangent function is x = arctan(y).
    Composite function properteis
    • tan(arctan(x)) = x.
    • arctan(tan(A)) = A.
    Example : Solve arctan(-1) = x.
    • By defintion tan(x) = -1.
    • Hence x = pi - pi/4. or x = 2*pi - pi/4.
    • Gerenal solution : x = n*pi - pi/4.
    Example : Solve arctan(+1) = x.
    • By defintion tan(x) = 1.
    • Hence x = pi/4. or x = pi + pi/4.
    • Gerenal solution : x = n*pi + pi/4.

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    TR 08 04. arcsin(x) + arcsin(y) = arcsin(x*Sqr(1-y^2) + y*Sqr(1-x^2))
    • Let A = arcsin(x) and then sin(A) = x. cos(A) = Sqr(1 - sin(A)^2) = Sqr(1 - x^2).
    • Let B = arcsin(y) and then sin(B) = y. cos(B) = Sqr(1 - sin(B)^2) = Sqr(1 - y^2).
    • Sin(A + B) = sin(A)*cos(B) + cos(A)*sin(B).
    • sin(A+B) = x*Sqr(1 - y^2) + y*Sqr(1 - x^2).
    • Hence A + B = arcsin(x*Sqr(1 - y^2) + y*Sqr(1 - x^2)).

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    Q05. arccos(x) + arccos(y) = arccos(x*y + Sqr(1-x^2)*Sqr(1-x^2))

    • Let A = arccos(x) and then cos(A) = x. sin(A) = Sqr(1 - cos(A)^2) = Sqr(1 - x^2).
    • Let B = arccos(y) and then cos(B) = y. sin(B) = Sqr(1 - cos(B)^2) = Sqr(1 - y^2).
    • cos(A + B) = cos(A)*cos(B) + sin(A)*sin(B).
    • cos(A + B) = x*y + Sqr(1-x^2)*Sqr(1 - y^2).
    • Hence A + B = arccos(x*y + Sqr(1 - x^2)*Sqr(1 - y^2)).

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    TR 08 06. arctan(x) + arctan(y) = arctan((x + y)/(1 - x*y))

    • Let A = arctan(x) and tan(A) = x.
    • Let B = arctan(y) and tan(B) = y.
    • tan(A + B) = (tan(A) + tan(B))/(1 - tan(A)*tan(B)).
    • tan(A + B) = (x + y)/(1 - x*y)
    • Hence A + B = arctan((x + y)/(1 - x*y))

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    TR 08 07. Express arccos(x) in terms of arctan(y)

    • In computer language, sometimes only give the arctan(x) as Atn(x).
    • To find arccos(x), we have to prepare a subroutine.
    • The method is given below
      • Use triangle method.
      • Let A = arccos(x) and then x = cos(A).
      • Since cos(A) = Adj/Hyp. Let Adj = x and Hyp = 1.
      • Hence Opp = Sqr(1 - x^2).
      • Hence tan(A) = Opp/Adj = Sqr(1 - x^2)/x.
      • Hence A = arctan(Sqr(1 - x^2)/x).
      • Hence arccos(x) = arctan(Sqr(1 - x^2)/x).

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    TR 08 08. arcsin(sin(A)+sin(B))+arcsin(sin(A)-sin(B))=pi/2, sin(A)^2+sin(B)^2 = ?
    Solution
    • Change arcsin(D) = d to sin(d) = D
      • Let arcsin(sin(A)+sin(B)) = x and sin(x) = sin(A) + sin(B)
      • Let arcsin(sin(A)-sin(B)) = y and sin(y) = sin(A) - sin(B)
    • Change sin(x)^2 + sin(y)^2 in terms of A and B
      • Change sin(x)^2 + sin(y)^2
      • = (sin(A) + sin(B))^2 + (sin(A) - sin(B))^2
      • = 2*(sin(A)^2 + sin(B)^2)
    • Also we have x + y = pi/2 or y = (pi/2) - x
      • sin(x)^2 + sin(y)^2
      • = sin(x)^2 + sin(pi/2 - x)
      • = sin(x)^2 + cos(x)^2
      • = 1
    • Hence (sin(A)^2 + sin(B)^2) = (sin(x)^2 + sin(y)^2)/2 = 1/2

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    TR 08 09. arcsin(4/5) + arcsin(5/13) + arcsin(16/65) = pi/2

    Proof
    • Let arcsin(4/5) = A then sin(A) = 4/5 and cos(A) = 3/5
    • Let arcsin(5/13) = B, then sin(B) = 5/13 and cos(B) = 12/13
    • Let arcsin(16/65) = C, then sin(C) = 16/65
    • Find cos(A + B)
      • cos(A + B) = cos(A)*cos(B) - sin(A)*sin(B)
      • = (3/5)*(12/13) - (4/5)*(5/13)
      • = (36 - 20)/65
      • = 16/65
      • = sin(C)
    • Since sin(C) = 16/65 and cos(pi/2 - C) = sin(C) = 16/65
    • cos(A + B) = cos(pi/2 - C)
    • Hence A + B + C = pi/2
    • Hence arcsin(4/5) + arcsin(5/13) + arcsin(16/65) = pi/2

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    TR 08 10. Prove that 2*arctan(1/3) + arctan(1/7) = 45 degrees

    Proof
    • Let arctan(1/3) = A and then tan(A) = 1/3
    • let arctan(1/7) = B and then tan(B) = 1/7
    • Tan(2*A) = 2*tan(A)/(1 - tan(A)^2)
    • = 2*(1/3)/(1 - (1/3)^2)
    • = 3/4
    • Tan(2*A + B) = (tan(2*A) + tan(B))/(1 - tan(2*A)*tan(B))
    • = (3/4 + 1/7)/(1 - (3/4)*(1/7))
    • = (21 + 4)/(28 - 3)
    • = 1
    • Hence 2*A + B = 45 degrees
    • Hence 2*arctan(1/3) + arctan(1/7) = 45 degrees

    Go to Begin

    TR 08 11. More arcsin(x) and arctan(x)
    Go to Begin

    TR 08 00.

    Properties of inverse functions
    • 1. arcsin(sin(A)) = A
    • 2. arccos(sin(A)) = A
    • 3. arctan(sin(A)) = A
    • 4. arccsc(sin(A)) = A
    • 5. arcsec(sin(A)) = A
    • 6. arccot(sin(A)) = A
    Properties of inverse functions
    • 1. sin(arcsin(x)) = x
    • 2. cos(arccos(x)) = x
    • 3. tan(arctan(x)) = x
    • 4. csc(arccsc(x)) = x
    • 5. sec(arcsec(x)) = x
    • 6. cot(arccot(x)) = x

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