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 . It’s inverse was a function with domain . 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:

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

Well, because arcsin is itself a one-to-one function, restricted to the domain 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 :

In fact, clearly there are an infinite number of solutions to the equation and we have just caught the one within the region . What are the other solutions. Well, we see that there are also solutions separated by . 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:

with .

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 ?

Let’s zoom into that one green point, and figure out how it relates to – the red point below:

So if this point can be written in terms of , then so can all the other points separated from the green point by intervals of . These are given by:

with .

So, altogether, the solutions to the equation are given by:

and with .

First year maths lecture notes subject links – MathemafricaMay 14, 2018 at 2:57 pm[…] Can we find the inverse of a function which is not one-to-one? – part 2 […]