Inverse Trigonometric Function Formulas:
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 inverse trigonometric formulas.
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^{-1}ϴ |
[–1, 1] |
[-\(\frac{π}{2}\), \(\frac{π}{2}\)] |
cos^{-1}ϴ |
[–1, 1] |
[0, π] |
tan^{-1}ϴ |
R |
[-\(\frac{π}{2}\), \(\frac{π}{2}\)] |
cot^{-1}ϴ |
R |
[0, π] |
sec^{-1}ϴ |
R – (–1, 1) |
[0, π] – {\(\frac{π}{2}\)} |
cosec^{-1}ϴ |
R – (–1, 1) |
[-\(\frac{π}{2}\), \(\frac{π}{2}\)] – {0} |
Properties of inverse trigonometric formula:
Functions and values |
Domain |
sin^{-1}\(\frac{1}{x}\) = cosec^{-1}x |
|x| ≥ 1 |
cos^{-1}\(\frac{1}{x}\) = sec^{-1}x |
|x| ≥ 1 |
tan^{-1}\(\frac{1}{x}\) = cot^{‑1}x |
X > 0 |
Some other inverse functions and their domain and range is given by:
Functions and values |
Domain |
sin^{–1}(–x) = – sin^{–1}x |
x ∈ [– 1, 1] |
tan–1 (–x) = – tan–1 x |
x ∈ R |
cosec^{–1}(–x) = – cosec^{–1}x |
|x| ≥ 1 |
Now cos, sec and cosec functions from Inverse trigonometric function are here.
Functions and values |
Domain |
cos^{–1}(–x) = π – cos^{–1}x |
x ∈ [– 1, 1] |
sec^{–1}(–x) = π – sec^{–1}x |
|x| ≥ 1 |
cot^{–1}(–x) = π – cot^{–1} x |
x ∈ R |
Identities of inverse trigonometric function formula:
Different identities are associated with Inverse trigonometric formula, Sincerely read and learn it.
Identities |
Domain |
sin^{-1}x + cos^{–1}x = \(\frac{π}{2}\) |
x ∈ [– 1, 1] |
tan^{–1}x + cot^{–1}x = \(\frac{π}{2}\) |
x ∈ R |
cosec^{–1}x + sec^{–1}x = \(\frac{π}{2}\) |
|x| ≥ 1 |
Common inverse trigonometric function formula:
Formulas |
Domain |
tan^{–1}x + tan^{–1}y = tan^{–1} \(\frac{x+y}{1-xy}\) |
xy < 1 |
tan^{–1}x + tan^{–1}y = tan^{–1} \(\frac{x-y}{1+xy}\) |
xy > -1 |
2 tan^{–1}x = tan^{-1}\(\frac{2x}{1-{x}^2}\) |
|x| < 1 |
All the conversion formula for inverse trigonometric functions are stated below. please take a look at all of these.
Formulas |
Domain |
2 tan^{–1}x = sin^{-1}\(\frac{2x}{1+{x}^2}\) |
|x| ≤ 1 |
2 tan^{–1}x = cos^{-1}\(\frac{1-{x}^2}{1+{x}^2}\) |
x ≥ 0 |
2 tan^{–1}x = tan^{-1}\(\frac{2x}{1-{x}^2}\) |
– 1 < x < 1 |
Some special inverse trigonometric function formula:
- sin^{-1}x + sin^{-1}y = sin^{-1} ( x\(\sqrt{1-{y}^2}\) + y\(\sqrt{1-{x}^2}\) )
if x, y ≥ 0 and x^{2}+ y^{2} ≤ 1.
- sin^{-1}x + sin^{-1}y = π – sin^{-1} ( x\(\sqrt{1-{y}^2}\) + y\(\sqrt{1-{x}^2}\) )
if x, y ≥ 0 and x^{2} + y^{2} > 1.
- sin^{-1}x – sin^{-1}y = sin^{-1} ( x\(\sqrt{1-{y}^2}\) – y\(\sqrt{1-{x}^2}\) )
if x, y ≥ 0 and x^{2}+ y^{2} ≤ 1.
- sin^{-1}x + sin^{-1}y = π – sin^{-1} ( x\(\sqrt{1-{y}^2}\) – y\(\sqrt{1-{x}^2}\) )
if x, y ≥ 0 and x^{2} + y^{2} > 1.
- cos^{−}^{1}x + cos^{−}^{1}y = cos^{−}^{1}(xy – \(\sqrt{1-{y}^2}\)\(\sqrt{1-{x}^2}\) )
if x, y ≥ 0 and x^{2}+ y^{2} ≤ 1.
- cos^{−}^{1}x + cos^{−}^{1}y = π – cos^{−}^{1}(xy – \(\sqrt{1-{y}^2}\)\(\sqrt{1-{x}^2}\) )
if x, y ≥ 0 and x^{2} + y^{2} > 1.
- cos^{−}^{1}x – cos^{−}^{1}y = cos^{−}^{1}(xy + \(\sqrt{1-{y}^2}\)\(\sqrt{1-{x}^2}\) )
if x, y ≥ 0 and x^{2}+ y^{2} ≤ 1.
- cos^{−}^{1}x – cos^{−}^{1}y = π – cos^{−}^{1}(xy + \(\sqrt{1-{y}^2}\)\(\sqrt{1-{x}^2}\) )
if x, y ≥ 0 and x^{2} + y^{2} > 1.
- tan^{−}^{1}x + tan^{−}^{1}y = tan^{−}^{1}(\(\frac{x+y}{1-xy}\) )
if x > 0, y > 0 and xy < 1.
- tan^{−}^{1}x + tan^{−}^{1}y = π + tan^{−}^{1}(\(\frac{x+y}{1-xy}\) )
if x > 0, y > 0 and xy > 1.
- tan^{−}^{1}x + tan^{−}^{1}y = tan^{−}^{1}(\(\frac{x+y}{1-xy}\) ) – π
if x < 0, y > 0 and xy > 1.
- tan^{−}^{1}x + tan^{−}^{1}y + tan^{−}^{1}z = tan^{−}^{1}(\(\frac{x+y+z-xyz}{1-xy-yz-zx}\) )
- tan^{−}^{1}x – tan^{−}^{1}y = tan^{−}^{1}(\(\frac{x-y}{1+xy}\) ) – π
if x < 0, y > 0 and xy > 1.
- 2 sin^{−}^{1}x = sin^{−}^{1} (2x\(\sqrt{1-{x}^2}\)
- 2 cos^{−}^{1}x = cos^{−}^{1} (2x^{2} – 1)
- 2tan^{-1}x = tan^{-1}( \(\frac{2x}{1-{x}^2}\) ) = sin^{-1}( \(\frac{2x}{1+{x}^2}\) ) = cos^{-1 }(\(\frac{1-{x}^2}{1+{x}^2}\))
- 3sin^{-1}x = sin^{-1}(3x – 4x^{3})
- 3 cos^{-1}x = cos^{-1}(4x^{3 }– 3x)
- 3tan^{-1}x = tan^{-1}( \(\frac{3x – {x}^3}{1 – 3{x}^2}\) )
Examples based on inverse trigonometric function formula:
- 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 \(\frac{π}{4}\).
- 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.
Inverse trigonometric formula here deals with all the essential trigonometric inverse function which will make it easy for you to learn anywhere and anytime. ITF formula for class 11 and 12 will help you in solving problems with needs.
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