Trigonometry is a part of geometry, where we will learn about the relationships between the angles and sides of a right-angled triangle. There are many trigonometry formulas and identities useful in mathematical and scientific calculations. There are many functions and ratios such as sin, cos, and tan. Similarly, we will have many inverse trigonometry concepts also. The inverse trigonometric functions are also useful. This article will explain the inverse trigonometry formula with examples. Let us learn the concept of it!

**What is Inverse Trigonometric Function?**

The inverse trigonometric functions are also popular as the anti-trigonometric functions. Sometimes these are also termed as arcus functions or cyclometric functions. The inverse trigonometric functions of various trigonometric ratios such as sine, cosine, tangent, cosecant, secant, and cotangent are defined.

These are useful to find the angle of a triangle from any of the known trigonometric functions. It is useful in many fields like geometry, engineering, physics, etc. To determine the sides of the triangle while the remaining side lengths are known, we may use these functions.

Source: en.wikipedia.org

The inverse functions are having the same name as the function but have the prefix word “arc”. Therefore, the inverse of sine will be arcsine, cosine will be arccosine, and tangent will be arctangent, similarly for others. But there is another notation is possible. Mostly arcsine is denoted by \(sin^{-1}\), arccosine is \(cos^{-1}\) and arctangent is \(tan^{-1}\). Thus,

\(\begin{align*}{\cos ^{ – 1}}\left( x \right) & = \arccos \left( x \right)\\ {\sin ^{ – 1}}\left( x \right) & = \arcsin \left( x \right)\\ {\tan ^{ – 1}}\left( x \right) & = \arctan \left( x \right)\end{align*}\)

If \(sin x = \frac{1}{2}\)

Then we may write it as: \(x = sin^{-1}\frac{1}{2}\)

This negative one must not be considered as the exponent -1.

So, we may differentiate as:

\({\left( {\cos \left( x \right)} \right)^{ – 1}} = \frac{1}{{\cos \left( x \right)}}\)

Thus inverse function will provide the value of angle from the value of trigonometric ratios.

**Some Formulae for Trigonometric Functions**

The inverse trigonometric formulae will help the students to solve the problems in an easy way with the application of these properties. Some of them are given here:

- \(sin^{-1}(-x) = -sin^{-1}x\)
- \(cos^{-1}(-x) = \pi – cos^{-1}x\)
- \(sin^{-1}(x) + cos^{-1}(x) = \frac{\pi}{2}\)
- \(tan^{-1}(x) + tan^{-1}(y) = \pi + tan^{-1} (\frac{x + y}{1 – xy})\)
- \(2 sin^{-1}(x) = sin^{-1}(2x\sqrt{1-x^{2}})\)
- \(3sin^{-1}(x) = sin^{-1}(3x-4x^{3})\)
- \(\sin ^{-1}x +\sin ^{-1}y=\sin ^{-1}(x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}), if x, y \geq 0 and x^{2}+y^{2} \leq 1\)
- \(\cos ^{-1}x +\cos ^{-1}y=\cos ^{-1}(xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}}) , if x, y >0 and x^{2}+y^{2} \leq 1\)

**Solved Examples for Inverse Trigonometry Formula**

Q.1: Find the values of \(sin (cos^{-1}\frac{3}{5})\) using inverse trigonometry formula.

Solution: Let us assume that, \(cos^{-1}\frac{3}{5} = \theta\)

Therefore, \(cos \theta = \frac{3}{5}\)

Therefore, \(sin \theta = \sqrt{(1 – cos^{2}} \theta)\)

\(sin \theta = \sqrt{(1 – \frac{9}{25} = \frac{4}{5}}\)

Thus, \(sin (cos^{-1}\frac{3}{5})= sin \theta = \frac {4}{5}.\)

Q.2: Evaluate \(\displaystyle {\cos ^{ – 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)\)

Solution: Let t = \(\displaystyle {\cos ^{ – 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)\)

\(\cos \left( t \right) = \frac{{\sqrt 3 }}{2}\)

For solving this, we have

\(\begin{align*} & \frac{\pi }{6} + 2\pi \,n\,,\;\;n = 0,\, \pm 1,\, \pm 2,\, \pm 3,\, \ldots \\ & \frac{{11\pi }}{6} + 2\pi \,n\,,\;\;n = 0,\, \pm 1,\, \pm 2,\, \pm 3,\, \ldots \end{align*}\)

Therefore, \({\cos ^{ – 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = \frac{\pi }{6}\)

I get a different answer for first example.

I got Q1 as 20.5

median 23 and

Q3 26

Hi

Same

yes