Question

How do you evaluate `tan(arcsin( 2/3))`?

Answer 1

The principal value gives

`tan text{Arc}text{sin}(2/3)= 2/sqrt{5}`

Treating `arcsin` as a multivalued expression gives

`tan arcsin(2/3) = pm 2/sqrt{5}`

Explanation:

This question has a different answer depending on whether we interpret `arcsin(2/3)` as all the angles whose sine is `2/3` or just the one in the first quadrant. I prefer the former interpretation, reserving `text{Arc}text{tan}(2/3)` for the principal value.

Let's answer first considering `arcsin(2/3)` to be a multivalued expression. That means

`theta = arcsin(2/3)`

is equivalent to

`sin theta = 2/3`

which is an equation with multiple solutions.

Then there are two possible values for `cos theta`:

`cos^2 theta + sin^2 theta = 1`

`cos theta = \pm \sqrt{1 - sin ^2 theta }`

In our case,

`cos theta = \pm sqrt{1 - (2/3)^2}= \pm sqrt{5}/3`

That's also apparent if we treat the right triangle in question as having opposite side `2` and hypotenuse `3` so other side `\sqrt{3^2-2^2}=sqrt{5}.`

So we get

`tan arcsin(2/3) = tan theta = {sin theta}/{cos theta} = {2/3}/{pm \sqrt{5}/3} = pm 2/sqrt{5}`

In the case when we're talking about the principal value of the inverse sine, a positive sine ends us in the first quadrant, so a positive tangent as well. We'll write this

`tan text{Arc}text{sin}(2/3)= 2/sqrt{5}`