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**What is an Inverse Function?**

f y=f(x) and x=g(y) are two functions such that f (g(y)) = y and g (f(y)) = x, then f and y are said to be inverse of each other i.e.,

g = f^{-1}

If y = f(x) then x = f^{-1} (y)

**Inverse Trigonometric Formulas**

Fig: Inverse Trigonometric Functions

The inverse trigonometric functions are the inverse functions of the trigonometric functions written as cos^{-1} x, sin^{-1} x, tan^{-1} x, cot^{-1} x, cosec^{-1} x, sec^{-1} x.

The inverse trigonometric functions are multivalued. For example, there are multiple values of ω such that z = sinω, so sin^{-1}z is not uniquely defined unless a principal value is defined.

Such principal values are sometimes denoted with a capital letter so, for example, the principal value of the inverse sine may be variously denoted sin^{-1}z or arcsinz.

Let’s say, if y = sin x , then x = sin^{-1} y, similarly for other trigonometric functions. This is one of the inverse trigonometric formulas. Now, y = sin^{-1} (x), y ∈ [π/2 , π/2] and x ∈ [-1,1].

- Thus, sin
^{-1}x has infinitely many values for given x ∈ [-1, 1]. - There is only one value among these values which lies in the interval [π/2, π/2]. This value is called the principal value.

**Domain and Range of Inverse Trigonometric Formulas**

Function | Domain | Range |

sin^{-1}x | [-1,1] | [-π/2,π/2] |

cos^{-1}x | [-1,1] | [0,π] |

tan^{-1}x | R | (-π/2,π/2) |

cot^{-1}x | R | (0,π) |

sec^{-1}x | R-(-1,1) | [0,π]-{π/2} |

cosec^{-1}x | R-(-1,1) | [-π/2,π/2]-{0} |

**Solved Examples**

**Ques 1: Find the exact value of each expression without a calculator, in [0,2π).**

- sin
^{-1}(−3√2) - cos
^{-1}(−2√2) - tan
^{-1}√3

**Ans:**

- Recall that −3√2 is from the 30−60−90 triangle. The reference angle for sin and 3√2 would be 60∘. Because this is sine and it is negative, it must be in the third or fourth quadrant. The answer is either 4π/3 or 5π/3.
- −2√2is from an isosceles right triangle. The reference angle is then 45∘. Because this is cosine and negative, the angle must be in either the second or third quadrant. The answer is either 3π/4 or 5π/4.
- √3is also from a 30−60−90 triangle. Tangent is √3 for the reference angle 60∘. Tangent is positive in the first and third quadrants, so the answer would be π/3 or 4π/3.

Notice how each one of these examples yields two answers. This poses a problem when finding a singular inverse for each of the trig functions. Therefore, we need to restrict the domain in which the inverses can be found.

**Ques 2: The value of 6 sin ^{-1}1**

Since sinπ/2=1,

6 sin

6 sin

**Ques 3: Find the value of tan ^{-1}(1.1106).Ans: **Let A=tan

Then, tanA = 1.1106

A = 48°

tan48 = 1.1106

[Use calculator in degree mode]

tan

**Properties of Inverse**

Here are the properties of the inverse trigonometric functions with proof.

**Property 1**

- sin
^{-1}(1/x) = cosec^{-1}x , x ≥ 1 or x ≤ -1 - cos
^{-1}(1/x) = sec^{-1}x , x ≥ 1 or x ≤ -1 - tan
^{-1}(1/x) = cot^{-1}x , x > 0

Proof : sin^{-1 }(1/x) = cosec^{-1}x , x ≥ 1 or x ≤ -1,

Let sin^{−1}x=y

i.e. x = cosec y

1/π = sin y

**Property 2**

- 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

Proof: sin^{-1}(-x) = -sin^{-1}(x), x ∈ [-1,1]

Let, Sin^{-1}(-x) = y

Then −x=sin y

x=−sin y

x=sin(−y)

sin^{−1}=sin^{−1}(sin(−y))

sin^{−1}x=y

sin^{−1}x=−sin^{−1}(−x)

Hence,sin^{−1}(−x)=−sin^{−1} x ∈ [-1,1]

**Property 3**

- 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

Proof : cos^{-1}(-x) = π – cos^{-1} x, x ∈ [-1,1]

Let cos^{−1}(−x)=y

cos y=−x x=−cos y

x=cos(π−y)

Since, cos π−q=−cos q

cos^{−1} x=π−y

cos^{−1} x=π–cos−1–x

Hence, cos^{−1}−x=π–cos^{−1}x

**Solved Example**

**Ques 1:** Prove that “sin^{-1}(-x) = – sin^{-1}(x), x ∈ [-1,1]”

**Ans: **Let, sin^{−1}(−x)=y

Then −x=sin y

x=−sin y

x=sin(−y)

sin^{−1} x=arcsin(sin(−y))

sin^{−1 }x=y

sin^{−1 }x=−sin^{−1}(−x)

Hence, sin^{−1}(−x)=−sin^{−1} x, x ∈ [-1,1]

This concludes our discussion on the topic of trigonometric inverse functions.**Ques 2: sin ^{-1}(cos π/3)=?Ans:** sin

= π/6 [substitute sin

**Ques 3: Find the value of sin (π/4+Cos ^{-1}(√2/2)).Ans: **

Let y=sin

Then, cosA= √2/2

Multiplying the numerator as well as denominator by √2 we get:

cosA=1/√2

A = π/4

Therefore

y = sin

y = sin (π/2)

hence, y=1.

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