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Mathematics Dictionary
Dr. K. G. Shih
Arctan(x)


  • Q01 | - Properties of arctan(x)
  • Q02 | - Find derivative of y = arctan(x)
  • Q03 | - Series of arctan(x)
  • Q04 | - Area under curve of y = 1/(1 + x^2)
  • Q05 | - Prove that arcsin(x) = arctan(x/Sqr(1 - x^2)
  • Q06 | - Formula

    • Q01. Properties of arctan(x)

    Deifintion of y = arctan(x)
    • If y = arctan(x) then x = tan(y)
    • Composite function
      • Arctan(tan(A)) = A.
      • Tan(arctan(x)) = x.
    Properties of Y=arctan(x)
      1. Domain : all real values of x
      2. Range : -pi/2 to pi/2
      3. The curve is always increasing since y' = 1/(1+x^2) > 0
      4. y" = -2*x/(1+x^2)^2 and point of inflection at x = 0
      5. Concavity
      • Concave upward if x < 0.
      • Concave downward if x > 0.
      6. Asymptote : y = pi/2 and y = -pi/2 when x = infinite
    Special values
    • arctan(0) = 0
    • arctan(Sqr(3)/3) = pi/6
    • arctan(1) = pi/4
    • arctan(Sqr(3)) = pi/3
    • arctan(+infinite) = pi/2
    Exercise
    • Draw the curve y = arctan(x) by using above properties.

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    Q02. Find derivative of y = arctan(x)

    • Let y = arctan(x) and then x = tan(y)
    • Take derivative on both sides
    • Hence 1 = (sec(y)^2)*y'
    • Hence y' = 1/(sec(y)^2)
    • Since 1 + tan(x)^2 = sec(x)^2
    • Hence y' = 1/(1 + tan(y)^2)
    • Hence y' = 1/(1+x^2)
    • Note : This the method to change trigonometry to algebra
    Change algebra to trigonometry
    • This is the method of the anti-derivative of y = arctan(x).

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    Q03. Series of arctan(x)
    Binomial Theory
    • (1 + x^2)^n = 1 + C(n,1)*(x^2) + C(n,2)*(x^2)^2 + ....
    • Let n = -1, then
      • C(n, 1) = n = -1
      • C(n, 2) = n*(n - 1)/(2!) = 1
      • C(n, 3) = n*(n - 1)*(n - 2) = -1
    • 1/(1 + x^2) = 1 - x + ((-1)*(-1 - 1)/(2!))*(x^4) + ....
    • = 1 - x^2 + x^4 - x^5 + .....
    Series of arctan(x)
    • Since ∫[1/(1 + x^2)]dx = arctan(x)
    • Using binomial theory we have
    • arctan(x) = ∫[1/(1 + x^2)]dx
    • = ∫[1 - x^2 + x^4 - ....]dx
    • = x - (x^3)/3 + (x^5)/5 - (x^7)/7 + .....
    Series of arctan(x) and pi
    • Arctan(1) = pi/4
    • Hence pi/4 = 1 - 1/3 + 1/5 - 1/7 + ....
    • This is very simple formula but it convergent very slow
    Go to Begin

    Q04. Area under curve of y = 1/(1 + x^2)

    Area under curve of y = 1/(1 + x^2)
    • From x = 0 to x = 1, the area is arctan(1) = pi/4
    • From x = 0 to x = infinte, the area is arctan(infinite) = pi/2
    Find area fro x = 0 to x = 2
    • Area = arctan(2) = ? (Use calculator)
    Go to Begin

    Q05. Prove that arcsin(x) = arctan(x/Sqr(1 - x^2)

    arcsin(x) = arctan(x/Sqr(1 - x^2)
    • Draw a right angle triangle
    • Let angle A = arcsin(x)
    • Then Opposite side is x
    • Hypothese is 1
    • Adjacent side = Sqr(1 - x^2)
    • Since tan(A) = Opp/Adj = x/Sqr(1 - x^2)
    • Hence A = arcsin(x) = arctan(x/sqr(1 - x^2))
    arccos(x) = arctan(Sqr(1 - x^2)
    • Draw a right angle triangle
    • Let angle A = arccos(x)
    • Then Adjacent side is x
    • Hypothese is 1
    • Opposite side = Sqr(1 - x^2)
    • Since tan(A) = Opp/Adj = Sqr(1 - x^2)/1
    • Hence A = arccos(x) = arctan(sqr(1 - x^2))
    Go to Begin

    Q06. Formula
    • tan(A) = Opp/Adj
    • 1 + tan(A)^2 = sec(A)^2
    • y = tan(x) and y' = sec(x)^2
    • y = arctan(x) and y' = 1/(1 + x^2)
    • ∫[1/(1+x^2)]dx = arctan(x)
    Values
    • tan(pi/4) = 1 and arctan(1) = pi/4
    • tan(pi/2) = ∝
    • tan(89.9999) = +∝
    • tan(90.0001) = -∝

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