Question

How do you evaluate `arcsin(sin((7pi)/3))`?

Answer 1

`arcsin(sin((7pi)/3))=(pi/3)`

Explanation:

If `arcsinx=theta`

then `sintheta=x`.

Let `theta=((7pi)/3)` and `sin((7pi)/3)=x`

Then `arcsin(sin((7pi)/3))=arcsinx=theta=(7pi)/3`

However, in case of trigonometric ratios, for each `x`, their could be multiple vales of `theta` (such as `2npi+theta`) e.g. not only `sin((7pi)/3)=x`, but so is for `sin(pi/3)`, `sin(-5pi/3)`, `sin(13pi/3)` etc.

Hence, the range for `theta=arcsinx` is limited to `[-pi/2,pi/2]`

Hence `arcsin(sin((7pi)/3))=(pi/3)`