The inverse trigonometric functions are arcus functions or anti trigonometric functions. These are the inverse functions of the trigonometric functions with suitably restricted domains. Here, we will study the inverse trigonometric formulae for the sine, cosine, tangent, cotangent, secant, and the cosecant functions, and are used to obtain an angle from any of the angle’s trigonometric ratios. Let us study them in detail.

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## What is an Inverse Function?

If 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**

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 multi-valued. 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 arc(sinz).

**Browse more Topics under Inverse Trigonometric Functions**

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 for You**

**Question 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

**Answer:**

- 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√2 is 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.
- √3 is 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.

**Question 2. The value of 6 sin ^{-1}1**

**Answer:** Let \( A=\sin^{-1} 1 \), then\( \sin A=1 \)

Since\( sin \pi /2=1 \),

\( 6\sin^{-1}1=6\times \frac{\pi }{2} \)

\( 6\sin^{-1} 1=3\pi \)

**Question 3. Find the value of tan ^{-1}(1.1106).**

**Answer:** Let \( A=\tan^{-1} (1.1106) \)

Then, \( \tan A=1.1106 \)

\( A=48° \)

\( \tan 48=1.1106 \)

[Use calculator in degree mode]

\( \tan^{-1} 1.1106=48° \)

### More Examples

**Question 4: sin ^{-1}(cos π/3)=?**

**Answer:** \( \sin^{-1} \left ( \cos \frac{\pi }{3} \right )=\sin^{-1} \frac{1}{2} \) [substitute cos(π/3)**=**1/2]

= \( \frac{\pi }{6} \) [substitute sin^{-1} (1/2) = π/6]

**Question 5: Find the value of sin (π/4+Cos ^{-1}(√2/2)).**

**Answer:** Let \( y=sin\left( \frac { π }{ 4 } +cos^{ -1 }\left( \frac { \sqrt { 2 } }{ 2 } \right) \right) \) and \( A=cos^{ -1 }\left( \frac { \sqrt { 2 } }{ 2 } \right) \)

Then, \(\cos A=\frac { \sqrt { 2 } }{ 2 } \)

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

\( \cos A=\frac { 1 }{ \sqrt { 2 } } \)

\( A=\frac{\pi }{4} \)

Therefore,

\( y=sin\left( \frac { π }{ 4 } +\frac { π }{ 4 } \right) \)

\( y=sin\left( \frac { π }{ 2 } \right)\)

hence, \( y=1 \)

**Question 6: Explain the inverse of cos?**

**Answer:** The arccos function happens to be the inverse of the cosine function. It returns the angle whose cosine happens to be a particular number.

**Question 7: Explain the inverse of sin?**

**Answer:** The inverse of the sin function happens to be the arcsin function. However, sine itself would not be invertible due to the fact that it’s not injective. Therefore, it’s certainly not bijective (invertible). To obtain the arcsine function, one will have to restrict the domain of sine to [−π2,π2].

**Question 8: Who is credited with the invention of inverse trigonometry? **

**Answer:** Daniel Bernoulli considered inverse trigonometric functions early in the 1700s. He used “A. sin” for a number’s inverse sine. Euler came forward with “A t” for the inverse tangent in the year 1736.

**Question 9: Explain what is meant by sec theta?**

**Answer:** The reciprocal cosine function happens to be secant: sec (theta)=1/cos(theta). The reciprocal sine function happens to be cosecant, csc(theta)=1/sin(theta). Also, the 3 new functions are cosecant theta, cotangent theta, and secant theta. Secant theta means 1 over x. Cotangent theta means x over y, and cosecant theta refers to 1 over y.

How to find domain of sin inverse3x-4xcube=3sin inverse x.

Is answer [-1,1]?

“For achieving the arcsine function we will be restricting the domain of ‘sin’ to ‘[−π2, π2]’”

in this solved examples question no. 4 have a mistake sin [-90 ,90] is domain of ‘sin’.

pleease correct these mistakes and many more mistakes also there in this web page content .

check these mistakes….