How to calculate simple trigonometric problem I tried for an hour or so to solve this but I can't show the way to the solution. How does one solve the below problem?
$\tan(\sin^{-1}(1/3))$?
Is the solution periodic because it is a tangent?
 A: Please look in the first row of the Wikipedia table on Relationships between trigonometric functions. There you find the relation that
$$\tan(\arcsin(x))=\frac{x}{\sqrt{1-x^2}}$$
This gives you immediately the correct result of 
$$\frac{1}{2\sqrt{2}}$$
Now, your task is to understand where this relation comes from.
As MichaelE2 pointed out, an alternative would be to ask WolframAlpha



A: To solve a problem like $\tan(\sin^{-1}(a/b))$, one can first set up a right triangle for $\sin^{-1}(a/b)$ by letting the hypotenuse be $b$ and the altitude be $a$. (If $a/b$ is negative, put the negative sign with $a$.) Then solve for the base, $\sqrt{b^2-a^2}$. The tangent is the altitude divided by the base, or $a / \sqrt{b^2-a^2}$.

A: Here is how you would do it if you didn't have Wikipedia.
$$\tan\left(\arcsin\left(\frac{1}{3}\right)\right)
=\frac{\sin\left(\arcsin\left(\frac{1}{3}\right)\right)}{\cos\left(\arcsin\left(\frac{1}{3}\right)\right)}
=\frac{\sin\left(\arcsin\left(\frac{1}{3}\right)\right)}{\sqrt{1-\sin^2\left(\arcsin\left(\frac{1}{3}\right)\right)}}
=\frac{\frac{1}{3}}{\sqrt{1-\left(\frac{1}{3}\right)^2}}=\frac{\sqrt{2}}{4}$$
