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:




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 [-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}:

singraphIn 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:


singraphp3So 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?