Inverse Trigonometric Functions

By Rajat Shukla |Updated : September 21st, 2016

Introduction

If sin x = 1/2, we can write one value of x = π/6.

If sin x = 1/3, i.e. x is not a well known angle, then we can write x = sin-1 1/3.

Similarly,

cos x = t ⇒ x = cos-1 t.

tan x = t ⇒ x = tan-1 t.

Rules for defined values of sin-1 x, cos-1 x

  1. y = sin-1 x:
    Domain: x Є [-1, 1]
    Range (principal value branch of sin-1 x)
    y Є [-π/2, π/2]

image002

  1. y = cos-1 x:
    Domain: x Є [-1, 1]
    Range (principal value branch of cos-1 x)
    y Є [0, π]

image004

  1. y = tan-1 x:
    Domain: x Є R
    Range (principal value branch of tan-1 x)
    y Є (-π/2, π/2)

image005

  1. y = cosec-1 x:
    Domain: x Є (-∞, -1] ∪ [1, ∞)
    Range (principal value branch of cosec-1 x)
    y Є [-π/2, 0) ∪ (0, π/2)

image006

  1. y = sec-1 x:
    Domain: x Є (-∞, -1] ∪ [l, ∞)
    Range (principal value branch of sec-1 x)
    y Є [0, π/2) ∪ (π/2, π]

image009

  1. y = cot-1 x:
    Domain: X Є R
    Range (principal value branch of cot-1 x)
    y Є (0, π)

Note the similarity in principal value branch of sin-1 x, cosec-1 x, tan-1 x.

Interval for allowed values of y is known as principal value branch of that inverse function.

Important Results

Important Results (I):

  1. sin (sin-1x) = x, cos (cos-1x) = x, ......
  2. sin-1 sin θ = θ, cos-1 cos θ = θ
    if θ allows the restrictions on y in the definition of corresponding inverse function.
    e.g. sin-1 sin2π/3 2π/3 because 2π/3 does not lie in the principal value branch of sin-1 x.
    Hence sin-1 sin2π/3 = sin-1 sin(π - π/3)sin-1 sinπ/3 = π/3.
  3. sin-1(-x) = -sin-1x cos-1(-x) = π - cos-1x
    cosec-1(-x) = -cosec-1x sec-1(-x) = π - sec-1x
    tan-1(-x) = -tan-1x cot-1(-x) = π - cot-1x

Important Results (II):

  1. If x > 0, y > 0 then
    image001
  2. lf x > 0, y > 0 then
    image002
  3. If x > 0
    image003
  4. image004

Important Results (III):

image005

 

Results (IV):

sin.-1 x + sin.-1 y= sin.-1 image006, when x ≥ 0, y ≥ 0, x2 + y2 ≤ 1

sin.-1 x + sin.-1 y= π - sin.-1 image006, when x ≥ 0, y ≥ 0, x2 + y2 > 1

Results (V):

sin.-1 x - sin.-1 y= sin.-1 image010, 0 ≤ y ≤ x

Results (VI):

cos.-1 x + cos.-1 y = cos.-1 image011, where x ≥ 0, y ≥ 0

cos.-1 x - cos.-1 y = cos.-1 image012, where 0 ≤ x, ≤ y

 

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