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

May 9, 2018

The principal value gives

$\tan \textrm{A r c} \textrm{\sin} \left(\frac{2}{3}\right) = \frac{2}{\sqrt{5}}$

Treating $\arcsin$ as a multivalued expression gives

$\tan \arcsin \left(\frac{2}{3}\right) = \pm \frac{2}{\sqrt{5}}$

#### Explanation:

This question has a different answer depending on whether we interpret $\arcsin \left(\frac{2}{3}\right)$ as all the angles whose sine is $\frac{2}{3}$ or just the one in the first quadrant. I prefer the former interpretation, reserving $\textrm{A r c} \textrm{\tan} \left(\frac{2}{3}\right)$ for the principal value.

Let's answer first considering $\arcsin \left(\frac{2}{3}\right)$ to be a multivalued expression. That means

$\theta = \arcsin \left(\frac{2}{3}\right)$

is equivalent to

$\sin \theta = \frac{2}{3}$

which is an equation with multiple solutions.

Then there are two possible values for $\cos \theta$:

${\cos}^{2} \theta + {\sin}^{2} \theta = 1$

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

In our case,

$\cos \theta = \setminus \pm \sqrt{1 - {\left(\frac{2}{3}\right)}^{2}} = \setminus \pm \frac{\sqrt{5}}{3}$

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

So we get

$\tan \arcsin \left(\frac{2}{3}\right) = \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{2}{3}}{\pm \setminus \frac{\sqrt{5}}{3}} = \pm \frac{2}{\sqrt{5}}$

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

$\tan \textrm{A r c} \textrm{\sin} \left(\frac{2}{3}\right) = \frac{2}{\sqrt{5}}$