Question

How do you find the `arcsin(sin((7pi)/6))`?

Answer 1

The answer is:
`arcsin(sin(7pi/6))=-pi/6`.

The range of a function `arcsin(x)` is, by definition ,
`-pi/2<=arcsin(x)<=pi/2`
It means that we have to find an angle `alpha` that lies between `-pi/2` and `pi/2` and whose `sin(alpha)` equals to a `sin(7pi/6)`.

From trigonometry we know that
`sin(phi+pi)=-sin(phi)`
for any angle `phi`.
This is easy to see if use the definition of a sine as an ordinate of the end of a radius in the unit circle that forms an angle `phi` with the X-axis (counterclockwise from the X-axis to a radius).
We also know that sine is an odd function, that is
`sin(-phi)=-sin(phi)`.

We will use both properties as follows:
`sin(7pi/6)=sin(pi/6+pi)=-sin(pi/6)=sin(-pi/6)`

As we see, the angle `alpha=-pi/6` fits our conditions. It is in the range from `-pi/2` to `pi/2` and its sine equals to `sin(7pi/6)`. Therefore, it's a correct answer to a problem.