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.

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!!!

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 : \[\begin{equation} e^{ix} = \cos x + i\sin x \end{equation}\] \[\begin{equation} e^{-ix} = \cos x - i\sin x \end{equation}\]

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.

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