So, in the last post we had seen that while the sin function is not one-to-one and thus doesn’t have an inverse, so long as we restrict it to a given domain, you will find that it is invertible. The domain that we found (indeed chose), was between $[-\frac{\pi}{2},\frac{\pi}{2}]$. It’s inverse was a function with domain $[-1,1]$. The name of the inverse is arcsin(x).  How can we use this to help us to solve problems?

Well, what if I asked you to solve:

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

You might think that because we have found the inverse of sin, that we can simply say that the solution to this is:

$x=arcsin\frac{1}{2}$

Well, because arcsin is itself a one-to-one function, restricted to the domain $[-1,1]$ this will clearly give us a single number (the answer is about 0.52):

Is that it then? Well, let’s look at the graph of sin(x) and see if this is the only solution to $sin(x)=\frac{1}{2}$:

In fact, clearly there are an infinite number of solutions to the equation $sin(x)=\frac{1}{2}$ and we have just caught the one within the region $[-\frac{pi}{2},\frac{\pi}{2}]$. What are the other solutions. Well, we see that there are also solutions separated by $2 \pi$. These are shown in the following plot (the original solution we found is the first red dot to the right of the y-axis):

These can be written as:

$\arcsin(\frac{1}{2})+2\pi k$ with $k\in \mathbb{Z}$.

However, it is clear that there are some more solutions that we haven’t captured.  How could we write the point shown in green in terms of $\arcsin(\frac{1}{2})$?

Let’s zoom into that one green point, and figure out how it relates to $\arcsin(\frac{1}{2})$ – the red point below:

So if this point can be written in terms of $arcsin(\frac{1}{2})$, then so can all the other points  separated from the green point by intervals of $2\pi$. These are given by:

$\pi-\arcsin(\frac{1}{2})+2\pi k$ with $k\in\mathbb{Z}$.

So, altogether, the solutions to the equation $sin x=\frac{1}{2}$ are given by:

$\arcsin(\frac{1}{2})+2\pi k$ and $\pi-\arcsin(\frac{1}{2})+2\pi k$ with $k\in\mathbb{Z}$.

 How clear is this post?