# Generalizing The Circular Functions

The basic idea behind this wiki is to demonstrate that how we can evaluate inverse trigonometric functions outside their domain using Complex Analysis And Euler's Formula.

#### Contents

## Introduction

Everyone of us knows that the range of circular functions : \(\sin x\) and \(\cos x\) is \([-1,1]\).

So did you ever tried solving the equation \[\sin x = 2\] or in particular, finding the solutions for : \[\sin x = n\ , \forall\ n \in \mathbb{R}\]

Yeah, you guessed it right, the solution is bit too complex!!!

## Derivation

### Basic Idea

Since, we are solving an equation, for such a point which is outside the range of a function, we know that there won't exist a real solution.

So, how should we start...?

In fact, because we're dealing with both complex numbers and trigonometric functions, that gives us a clue of starting with the Euler's Identity : \(e^{ix} = \cos x + i\sin x\)

\[\] \[\]

### Method

Today, I am going to introduce you to a method with which you can easily evaluate \(\arccos (x)\) and \(\arcsin (x)\) for all real value of \(x\).

So, here we go -

First of all, by Euler Formula, we have : \[\begin{equation} e^{ix} = \cos x + i\sin x \end{equation}\] \[\begin{equation} e^{-ix} = \cos x - i\sin x \end{equation}\]

Subtracting them up, gives \[\sin x = \frac{e^{ix}-e^{-ix}}{2i}\]

Now, we wish to find the solutions for \(\sin x = n\). So, \[n = \frac{e^{ix}-\frac{1}{e^{ix}}}{2i}\]

So, now can you look up the quadratic coming around...?

No! ok, I'll just a use a simple substitution here, which makes the work tidier and easier to see the quadratic.

Substituting \(e^{ix} = t\), we get \[n = \frac{t-\frac{1}{t}}{2i}\]

Upon Rearranging, we get \[\Rightarrow t^2 - 2int - 1 = 0\]

Now, that's a quadratic in \(t\) whose solutions are - \[t = e^{ix} = in\pm\sqrt{1-n^2}\]

Taking natural logarithm and multiplying by \(-i\) on both \(L.H.S\) and \(R.H.S\) yields - \[x = -i\ln{\left(in\pm\sqrt{1-n^2}\right)}\] And thus, \[\boxed{\arcsin(n) = -i\ln{\left(in\pm\sqrt{1-n^2}\right)}}\]

Since, \[\frac{-\pi}{2} \leq \arcsin(x) \leq \frac{\pi}{2}\] so \(\arcsin(x)\) lies in the first and fourth quadrants, and hence we do not need to make any changes in our formula.

## Practice Problems

1. Evaluate : \[\arcsin(2015)\] 2. Solve for \(x\) : \[\cos x = n\ ,\ \forall\ n \in \mathbb{R}\]

**Cite as:**Generalizing The Circular Functions.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/generalizing-the-circular-functions/