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Inverse Trigonometric Function Formula, Identities, Examples, Domain, Range

inverse trigonometric function formula

Also, read
Basics of Trigonometry
Trigonometry for class 10
Trigonometric Formulas
Applications of Trigonometry

Inverse Trigonometric Function:

While studying calculus we see that Inverse trigonometric function plays a very important role. For every section of trigonometry with limited inputs in function, we use inverse trigonometric function formula to solve various types of problems. In this article, we have listed all the important formulas and identities related to inverse trigonometric functions.

Domain and range of simple trigonometric functions:

Functions Domain Range
sinϴ R [-1, 1]
cosϴ R [-1, 1]
tanϴ R- { x : x = (2n + 1)\(\frac{π}{2}\), n ∈ Z} R
cotϴ R – { x : x = nπ, n ∈ Z} R
secϴ R- { x : x = (2n + 1)\(\frac{π}{2}\), n ∈ Z} R – (-1, 1)
cosecϴ R – { x : x = nπ, n ∈ Z} R –(-1, 1)

Domain and range of inverse trigonometric function formula:

Functions Domain Range(principal value)
sin-1ϴ [–1, 1] [-\(\frac{π}{2}\), \(\frac{π}{2}\)]
cos-1ϴ [–1, 1] [0, π]
tan-1ϴ R [-\(\frac{π}{2}\), \(\frac{π}{2}\)]
cot-1ϴ R [0, π]
sec-1ϴ R – (–1, 1) [0, π] – {\(\frac{π}{2}\)}
cosec-1ϴ R – (–1, 1) [-\(\frac{π}{2}\), \(\frac{π}{2}\)] – {0}

Properties of inverse trigonometric function formula:

Functions and values Domain
sin-1\(\frac{1}{x}\) = cosec-1x |x| ≥ 1
cos-1\(\frac{1}{x}\) = sec-1x |x| ≥ 1
tan-1\(\frac{1}{x}\) = cot‑1x X > 0

 

Functions and values Domain
sin–1(–x) = – sin–1x x ∈ [– 1, 1]
tan–1 (–x) = – tan–1 x x ∈ R
cosec–1(–x) = – cosec–1x |x| ≥ 1

 

Functions and values Domain
cos–1(–x) = π – cos–1x x ∈ [– 1, 1]
sec–1(–x) = π – sec–1x |x| ≥ 1
cot–1(–x) = π – cot–1 x x ∈ R

Identities of inverse trigonometric function formula:

Identities Domain
sin-1x + cos–1x = \(\frac{π}{2}\) x ∈ [– 1, 1]
tan–1x + cot–1x = \(\frac{π}{2}\) x ∈ R
cosec–1x + sec–1x = \(\frac{π}{2}\) |x| ≥ 1

Common inverse trigonometric function formula:

Formulas Domain
tan–1x + tan–1y = tan–1 \(\frac{x+y}{1-xy}\) xy < 1
tan–1x + tan–1y = tan–1 \(\frac{x-y}{1+xy}\) xy > -1
2 tan–1x = tan-1\(\frac{2x}{1-{x}^2}\) |x| < 1

 

Formulas Domain
2 tan–1x = sin-1\(\frac{2x}{1+{x}^2}\) |x| ≤ 1
2 tan–1x = cos-1\(\frac{1-{x}^2}{1+{x}^2}\) x ≥ 0
2 tan–1x = tan-1\(\frac{2x}{1-{x}^2}\) – 1 < x < 1

Some special inverse trigonometric function formula:

  1. sin-1x + sin-1y = sin-1 ( x\(\sqrt{1-{y}^2}\) + y\(\sqrt{1-{x}^2}\) )
    if x, y ≥ 0 and x2+ y2 ≤ 1.
  2. sin-1x + sin-1y = π – sin-1 ( x\(\sqrt{1-{y}^2}\) + y\(\sqrt{1-{x}^2}\) )
    if x, y ≥ 0 and x2  + y2 > 1.
  3. sin-1x – sin-1y = sin-1 ( x\(\sqrt{1-{y}^2}\) – y\(\sqrt{1-{x}^2}\) )
    if x, y ≥ 0 and x2+ y2 ≤ 1.
  4. sin-1x + sin-1y = π – sin-1 ( x\(\sqrt{1-{y}^2}\) – y\(\sqrt{1-{x}^2}\) )
    if x, y ≥ 0 and x2  + y2 > 1.
  5. cos1x + cos1y = cos1(xy – \(\sqrt{1-{y}^2}\)\(\sqrt{1-{x}^2}\) )
    if x, y ≥ 0 and x2+ y2 ≤ 1.
  6. cos1x + cos1y = π – cos1(xy – \(\sqrt{1-{y}^2}\)\(\sqrt{1-{x}^2}\) )
    if x, y ≥ 0 and x2  + y2 > 1.
  7. cos1x – cos1y = cos1(xy + \(\sqrt{1-{y}^2}\)\(\sqrt{1-{x}^2}\) )
    if x, y ≥ 0 and x2+ y2 ≤ 1.
  8. cos1x – cos1y = π – cos1(xy + \(\sqrt{1-{y}^2}\)\(\sqrt{1-{x}^2}\) )
    if x, y ≥ 0 and x2  + y2 > 1.
  9. tan1x + tan1y = tan1(\(\frac{x+y}{1-xy}\) )
    if x > 0, y > 0 and xy < 1.
  10. tan1x + tan1y = π + tan1(\(\frac{x+y}{1-xy}\) )
    if x > 0, y > 0 and xy > 1.
  11. tan1x + tan1y = tan1(\(\frac{x+y}{1-xy}\) ) – π
    if x < 0, y > 0 and xy > 1.
  12. tan1x + tan1y + tan1z = tan1(\(\frac{x+y+z-xyz}{1-xy-yz-zx}\) )
  13. tan1x – tan1y = tan1(\(\frac{x-y}{1+xy}\) ) – π
    if x < 0, y > 0 and xy > 1.
  14. 2 sin1x = sin1 (2x\(\sqrt{1-{x}^2}\)
  15. 2 cos1x = cos1 (2x2 – 1)
  16. 2tan-1x = tan-1( \(\frac{2x}{1-{x}^2}\) ) = sin-1( \(\frac{2x}{1+{x}^2}\) ) = cos-1 (\(\frac{1-{x}^2}{1+{x}^2}\))
  17. 3sin-1x = sin-1(3x – 4x3)
  18. 3 cos-1x = cos-1(4x3 – 3x)
  19. 3tan-1x = tan-1( \(\frac{3x – {x}^3}{1 – 3{x}^2}\) )

Examples based on inverse trigonometric function formula:

  1. Find the principal value of sin–1(\(\frac{1}{\sqrt{2}}\) ).
    Solution:
    Let sin–1(\(\frac{1}{\sqrt{2}}\) ) = y. Then, sin y = ( \(\frac{1}{\sqrt{2}}\) )

    We know that the range of the principal value branch of sin–1 is [-\(\frac{π}{2}\), \(\frac{π}{2}\)].
    Also, sin( \(\frac{π}{4}\) ) = \(\frac{1}{\sqrt{2}}\)
    so, principal value of sin-1( \(\frac{1}{\sqrt{2}}\) ) is \(\frac{π}{4}\).
  2. Write cot-1( \(\frac{1}{\sqrt{{x}^2-1}}\) ), | x | > 1 in the simplest form.
    Solution:
    Let x = sec θ, then \(\sqrt{{x}^2-1}\) = \(\sqrt{{sec}^2θ-1}\) = tan θ

    Therefore, cot–1= \(\frac{1}{\sqrt{x^{2}–1}}\) = cot–1 (cot θ) = θ = sec–1 x, which is the simplest form.

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