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Inverse Trigonometric Functions are the inverse functions of some basic trigonometric functions, such as sine, cosine, tangent, secant, cosecant, and cotangent. Inverse trigonometric functions have qualities that are determined by the domain and range of the functions. Anti trigonometric functions, arcus functions, and cyclometric functions are all examples of inverse trigonometric functions. The formula of these functions helps us to determine any angles with the help of any trigonometric ratios. In this article, we will have a look at the inverse trigonometric functions, their properties, and some solved examples.
Read More: Applications of Trigonometry
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Key Terms: Inverse Trigonometric Functions, Angles, Perpendicular, sine, cosine, tangent, secant, cosecant, cotangent, trigonometric function
What are Inverse Trigonometric Functions?
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Inverse trigonometric functions are frequently referred to as "Arc Functions." They create the length of arc required to obtain a given value of a trigonometric function for a given value. The range of values that an inverse function can achieve with the defined domain of the function is defined as the range. A function's domain is defined as the collection of all conceivable independent variables in which the function can exist. Inverse trigonometric functions have a fixed range of values. These inverse trigonometric functions formulas allow us to find any angle with any Trigonometry Ratio. The properties of trigonometric functions are used to create these formulas.
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Domain and Range of Inverse & Inverse Trigonometric Functions
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With respect to the domain and range of a function, the following formulas are important:
Inverse Functions: Domain and Range
- sin(sin−1 x) = x, if -1 ≤ x ≤ 1
- cos(cos−1x) = x, if -1 ≤ x ≤ 1
- tan(tan−1x) = x, if -∞ ≤ x ≤∞
- cot(cot−1x) = x, if -∞≤ x ≤∞
- sec(sec−1x) = x, if -∞ ≤ x ≤ -1 or 1 ≤ x ≤ ∞
- cosec(cosec−1x) = x, if 1 ≤ x ≤ ∞ or -∞ ≤ x ≤ -1
Inverse Trigonometric Functions: Domain and Range
In case of Inverse Trigonometric Functions, the formulas are:
- sin−1(sin y) = y, if -π/2 ≤ y ≤ π/2
- cos−1(cos y) = y, if 0 ≤ y ≤ π
- tan−1(tan y) = y, if -π/2 < y < π/2
- cot−1(cot y) = y if 0 < y < π
- sec−1(sec y) = y, if 0 ≤ y ≤ π, y ≠ π/2
- cosec−1(cosec y) = y, if -π/2 ≤ y ≤ π/2, y ≠ 0
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Important Properties of Inverse Trigonometric Functions
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We will learn about the properties of six inverse trigonometric functions in this section. These properties are considered for simplifying expressions and solving equations using inverse trigonometric functions.
Property 1
- sin−1(x) = cosec–1(1/x) θ for all x ∈ [-1, 1] −{0}
- cos−1(x) = sec–1(1/x) θ for all x ∈ [-1, 1] −{0}
- tan−1(x) = cot−1 (1/x) − π, if x < 0 OR cot–1(1/x) for x > 0
- cot−1(x) = tan–1(1/x) + π for x < 0 OR tan–1(1/x), if x > 0
Property 2
- sin (sin−1 x) = x for all x ∈ [-1, 1]
- cos (cos−1 x) = x for all x ∈ [-1, 1]
- tan (tan−1 x) = x for all x ∈ R
- cosec (cosec−1 x) = x for all x ∈ (∞, -1] ∪ [1, ∞)
- sec (sec−1 x) = x for all x ∈ (-∞, -1] ∪ [1, ∞)
- cot (cot−1 x) = x for all x ∈ R
Property 3
- sin−1 (\(\frac{1}{x}\)) = cosec-1 x for all x ∈ (∞, -1] ∪ [1, ∞)
- cos−1 (\(\frac{1}{x}\)) = sec-1 x for all x ∈ (-∞, -1] ∪ [1, ∞)
- tan−1 (\(\frac{1}{x}\)) = cot-1 x, for x > 0 and -π + cot−1 x, for x < 0
Proof: Sin−1 (1/x) = cosec−1x, for x ≥ 1 or x ≤ −1
Let cosec-1x =y, i.e. x= cosec y
⇒ (1/x)= sin y
Therefore, sin−1(1/x)= y
Or, sin-1(1/x) = cosec-1x
In the same way, we can obtain the other results.
Property 4
- sin-1 (-x) = - sin-1 x for all x ∈ [-1, 1]
- cos-1(-x) = π - cos-1 x for all x ∈ [-1, 1]
- tan-1 (-x) = - tan-1 x for all x ∈ R
- cosec-1 (-x) = - cosec-1 x for all x ∈ (∞, -1] ∪ [1, ∞)
- sec-1 (-x) = π - sec-1 x for all x ∈ (-∞, -1] ∪ [1, ∞)
- cot-1 (-x) = π - cot-1 x for all x ∈ R
Proof:-
- Cos−1 (−x) = π − Cos−1(x)
Let cos−1(−x) = y
i.e., − x = cos y
⇒ x= −cos y
= cos(π–y)
Therefore, cos-1(x) = π–y
Or, cos-1(x) = π – cos-1(−x)
Therefore, cos-1(−x) = π – cos-1(x)
In the same way, we can obtain the following results:
- sec-1(−x) = π – sec-1x, for |x| ≥ 1
- cot-1(−x) = π – cot-1x, for x ∈ R
- Sin−1 (−x) = − Sin−1(x)
Let sin-1(−x)= y,
i.e.,−x= sin y
We get, x = − sin y
Thus, x = sin (−y)
Or, sin-1(x) = − y
= − sin-1(−x)
Therefore, sin−1(−x) = − sin−1(x)
In the same way, we can obtain the following results:
- cosec-1(−x) = − cosec-1x, for |x| ≥ 1
- tan-1(−x) = − tan-1x, for x ∈ R
Property 5
- sin-1 x + cos-1 x = π/2 for all x ∈ [-1, 1]
- tan-1 x + cot-1 x = π/2 for all x ∈ R
- sec-1 x + cosec-1 x = π/2 for all x ∈ (-∞, -1] ∪ [1, ∞)
Proof: sin−1(x) + cos−1 (x) = (π/2), for x ∈ [−1,1]
Let sin−1(x)= y,
i.e., x= sin y
= cos((π/2)−y)
⇒ cos−1(x) = (π/2)–y
= (π/2) − sin−1 (x)
Therefore, sin−1(x) + cos−1(x)= (π/2)
In the same way, we can obtain the following results:
- tan−1(x) + cot−1(x) = (π/2), for x ∈ R
- cosec−1(x) + sec−1(x) = (π/2), for |x| ≥ 1
Read More: Addition And Subtraction Of Integers
Property 6
If x and y are greater than 0
- tan−1 x + tan−1 y = π + tan−1 (\(\frac{x + y}{1 - xy}\)), if xy > 1
- tan−1 x + tan−1 y = tan−1 (\(\frac{x + y}{1 - xy}\)), if xy < 1
If x and y are less than 0
- tan−1 x + tan−1 y = tan−1 (\(\frac{x + y}{1 - xy}\)), if xy < 1
- tan−1 x + tan−1 y = - π + tan−1 (\(\frac{x + y}{1 - xy}\)), xy < 1
For example:
-
Tan−1(−2) + Tan−1 (−3) = Tan−1 [(−2 + −3)/(1−6)]
= Tan−1(−5/−5) = Tan−11
= π/4
-
Tan−1(−½) + Tan−1(−1/3) = Tan−1[(−½ − 1/3)/(1− 1/6)]
= tan−1(−1)
= −π/4
- sin−1(cos θ) = π/2 − θ, for θ ∈ [0,π]
- cos−1(sin θ) = π/2 − θ, for θ ∈ [−π/2 , π/2]
- tan−1(cot θ) = π/2 − θ, for θ ∈ [0,π]
- cot−1(tan θ) = π/2 − θ, for θ ∈ [−π/2, π/2]
- sec−1(cosec θ) = π/2 − θ, for θ ∈ [−π/2, 0] ∪ [0, π/2]
- cosec−1(sec θ) = π/2 − θ, for θ ∈ [0,π] − {π/2}
- sin−1(x) = cos−1 [√(1−x2)], for 0≤x≤1
= −cos−1 [√(1−x2)], for −1≤x<0
Property 8
- sin−1 x + sin−1 y = sin−1 { x \(\sqrt{(1- y^2)}\) + y \(\sqrt{(1- x^2)}\) }
- cos−1 x + cos−1 y = cos−1 { xy - \(\sqrt{(1- x^2)}\) \(\sqrt{(1- y^2)}\) }
Property 9
- 2 sin-1 x = sin-1 (2x\(\sqrt{(1- x^2)}\)), if \(\frac{-1}{\sqrt {2}}\) ≤ x ≤ \(\frac{1}{\sqrt {2}}\)
= π - sin-1 (2x \(\sqrt{(1- x^2)}\)), if \(\frac{1}{\sqrt {2}}\) ≤ x ≤ 1
= - π - sin−1 (2x \(\sqrt{(1- x^2)}\)), if -1 ≤ x ≤ \(\frac{-1}{\sqrt {2}}\)
- 3 sin-1 x = sin−1 (3x - 4x3), if \(\frac{-1}{2}\) ≤ x ≤ \(\frac{1}{2}\)
= π - sin−1 (3x - 4x3), if \(\frac{1}{2}\) < x ≤ 1
= - π - sin−1 (3x - 4x3), if -1 ≤ x < \(\frac{-1}{2}\)
- 2 cos−1 x = cos-1 (2x2 - 1), if 0 ≤ x ≤ 1
= 2 π - cos−1 (2x2 - 1), if -1 ≤ x ≤ 0
= cos−1 (4x3 - 3x), if \(\frac{1}{2}\) ≤ x ≤ 1
- 3 cos-1 x = 2 π + cos-1 (2x2 - 1), if \(\frac{-1}{2}\) ≤ x ≤ \(\frac{1}{2}\)
= - 2 π + cos-1 (2x2 - 1), if if -1 ≤ x ≤ \(\frac{-1}{2}\)
- 2 tan-1 x = tan−1 (\(\frac{2x}{1 - x^2}\)), if -1 < x < 1
= π + tan−1 (\(\frac{2x}{1 - x^2}\)), if x > 1
= - π + tan−1 (\(\frac{2x}{1 - x^2}\)), if x < -1
- 3 tan-1 x = tan−1 (\(\frac{3x-x^3}{1 -3 x^2}\)), if \(\frac{-1}{\sqrt {3}}\) < x < \(\frac{1}{\sqrt {3}}\)
= π + tan−1 (\(\frac{3x-x^3}{1 -3 x^2}\)), if x > \(\frac{1}{\sqrt {3}}\)
= - π + tan−1 (\(\frac{3x-x^3}{1 -3 x^2}\)), if x <\(\frac{-1}{\sqrt {3}}\)
- 2 tan-1 x = sin-1 (\(\frac{2x}{1 + x^2}\)), if -1 ≤ x ≤ 1
= π - sin−1 (\(\frac{2x}{1 + x^2}\)), if x > 1
= - π - sin−1 (\(\frac{2x}{1 + x^2}\)), if x < -1
- 2 tan-1 x = cos-1 (\(\frac{1- x^2}{1 + x^2}\)), if 0 ≤ x < ∞
= - cos−1 (\(\frac{1- x^2}{1 + x^2}\)), if -∞ < x ≤ 0
Also Read:
Things to Remember
- → "Arc Functions" is the other name for inverse trigonometric functions. They create the length of arc required to obtain a given value of a trigonometric function for a given value.
- → The range of an inverse function is defined as the set of inverse function values that can be obtained within the function's defined domain.
- → A function's domain is defined as the collection of all conceivable independent variables in which the function can exist.
- → With any of the Trigonometry ratios, inverse trigonometric functions are applied to compute the unknown values of angles in a right-angle triangle.
- → Each of the trigonometry ratios has six trigonometric functions. Arcsine, Arccosine, Arctangent, Arcsecant, Arcosecant, and Arccotangent are the inverses of those six trigonometric functions.
Read More: Trigonometry Ratio Important Formula
Sample Questions
Ques. Solve for x: 2 tan-1x (cos x) = tan-1(2 cosec x) (3 marks)
Ans. 2 tan−1x (cos x) = tan−1(2cosec x)
tan−1 (\(\frac{2 cos x}{1- cos^2x}\)) = tan−1 (2cosec x)
\(\frac{2 cos x}{sin^2x}\)= 2 cosec x
cos x = sin x
tan x = 1
tan x = tan (π/4)
x = nπ + (π/4), n ∈ Z (Ans.)
Ques. Evaluate tan (2 tan-1 \(\frac{1}{5}\)) (2 marks)
Ans. tan (2 tan-1 \(\frac{1}{5}\))
= tan {tan−1( \(\frac{2 \times \frac{1}{5}}{1- \frac{1}{25}}\) )}
= tan (tan−1 \(\frac{5}{12}\))
= \(\frac{5}{12}\) (Ans.)
Ques. Prove that cos-1 \(\frac{12}{13}\)+ sin-1 \(\frac{3}{5}\) = sin-1 \(\frac{56}{65}\). (3 marks)
Ans: cos-1 \(\frac{12}{13}\)+ sin-1 \(\frac{3}{5}\)= sin-1 \(\frac{5}{13}\)+ sin-1 \(\frac{3}{5}\)(As cos-1 \(\frac{12}{13}\)= sin-1 \(\frac{5}{13}\) )
= sin-1 {\(\frac{5}{13}\)* \(\sqrt{1-( 3/5)^2}\)+\(\frac{3}{5}\) * \(\sqrt{1-( 5/13)^2}\) }
= sin-1 {\(\frac{5}{13}\)* \(\frac{4}{5}\)+ \(\frac{3}{5}\)* \(\frac{12}{13}\)} = sin-1 \(\frac{56}{65}\)
Ques. If sin (cos-1 (5/13) + sin-1 x) = 1, find the value of x. (3 marks)
Ans. sin (cos-1 (5/13) + sin-1 x) = 1
cos-1 (5/13) + sin-1 x = π/2
sin-1 x = (π/2) - cos-1 (5/13) (As sin-1 (5/13) + cos-1 (5/13) = π/2)
sin-1 x = sin-1 (5/13)
x = 5/13
Ques. Prove: sin-1 (-x) = - sin-1 (x), x ∈ (-1, 1) (3 marks)
Ans: Let Sin-1 (-x) = y
So, -x = sin y
x = - sin y
x =sin (-y)
sin-1 -x = arcsin ( sin(-y))
sin-1 -x = y
So, Sin-1 (-x) = - Sin-1 (x), x ∈ (-1, 1)
Ques. Solve x: cos (tan-1 x) = sin (cot-1 (3/4)) (5 marks)
Ans. As we have cos (tan-1 x) = sin (cot-1 (3/4))
So, cos (cos-1 \(\frac{1}{\sqrt{1 + x^2}}\)) = sin (sin-1 (4/5))
\(\frac{1}{\sqrt{1 + x^2}}\) = (4/5)
4 \(\sqrt{1 + x^2}\) = 5
16 (\(\sqrt{1 + x^2}\)) = 25
16x2 = 9
x = ± (3/4)
Ques. Find sin (cos-1 3/5) (3 marks)
Ans. Consider cos-1 3/5 = x
So, cos x will be = 3/5
We also know that, sin x = \(\sqrt{1- cos^2x}\)
Accordingly, sin x = \(\sqrt{1-(9/25)}\) = 4/5
Hence sin x will be = sin (cos-1 3/5) = 4/5
Ques. Prove -cos-1 (4x3 -3 x) =3 cos-1 x , ½ ≤ x ≤ 1. (5 marks)
Ans. Consider x = cos θ, where θ = cos-1 (-x)
On LHS = cos-1 (-x) (4x3 -3x)
Substituting the value of x, we find that
= cos-1 (-x) (4 cos3θ - 3 cosθ)
According to question, we also find that cos-1 (cos 3θ) = 3θ
By substituting the value of θ, we get
= 3 cos-1 x = RHS
LHS= RHS, hence proved.
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