# How do you evaluate  tan (arcsin (3/4))?

Apr 12, 2018

$\tan \left(\arcsin \left(\frac{3}{4}\right)\right) = \frac{3 \sqrt{7}}{7}$

#### Explanation:

Note that since $\frac{3}{4} > 0$, the angle $\theta = \arcsin \left(\frac{3}{4}\right)$ will be in Q1. So we can take it to be an acute angle of a right angled triangle.

Remember:

$\sin \theta = \text{opposite"/"hypotenuse}$

$\tan \theta = \text{opposite"/"adjacent}$

So consider a right angled triangle with hypotenuse $4$, one leg $3$ and the other $\sqrt{{4}^{2} - {3}^{2}} = \sqrt{16 - 9} = \sqrt{7}$. This last leg is the one adjacent to $\theta$.

So:

$\tan \left(\arcsin \left(\frac{3}{4}\right)\right) = \text{opposite"/"adjacent} = \frac{3}{\sqrt{7}} = \frac{3 \sqrt{7}}{7}$

Alternatively, using:

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

we find:

$\cos \left(\theta\right) = \pm \sqrt{1 - {\sin}^{2} \theta} = \pm \sqrt{1 - {\left(\frac{3}{4}\right)}^{2}} \pm \sqrt{1 - \frac{9}{16}} = \pm \frac{\sqrt{7}}{4}$

Then since $\arcsin \left(\theta\right) \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right]$, where $\cos > 0$, we require the $+$ sign and find:

$\cos \left(\theta\right) = \frac{\sqrt{7}}{4}$

Then:

$\tan \left(\arcsin \left(\frac{3}{4}\right)\right) = \tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right) = \frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}} = \frac{3}{\sqrt{7}} = \frac{3 \sqrt{7}}{7}$