# 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 : $e^{ix} = \cos x + i\sin x$

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### 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 : $e^{ix} = \cos x + i\sin x$ $e^{-ix} = \cos x - i\sin x$

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/