# 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?

-
Solve for what? To solve something you need at least an equation. You have only an expression. Do you search the roots, but if yes, where is your variable? –  halirutan Nov 2 '13 at 12:28
@halirutan This is a standard mathematics problem in trigonometry. By "solve", the OP means evaluate. He needs to set up a right triangle and work it out. It is not a question about the software Mathematica. –  Michael E2 Nov 2 '13 at 12:32

## migrated from mathematica.stackexchange.comNov 2 '13 at 13:26

This question came from our site for users of Mathematica.

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

-
–  Michael E2 Nov 2 '13 at 12:47
$$\tan\left(\arcsin\left(\frac{1}{3}\right)\right) =\frac{\sin\left(\arcsin\left(\frac{1}{3}\right)\right)}{\cos\left(\arcsin\left(\frac{1}{3}\right)\right)} =\frac{\sin\left(\arcsin\left(\frac{1}{3}\right)\right)}{\sqrt{1-\sin^2\left(\arcsin\left(\frac{1}{3}\right)\right)}} =\frac{\frac{1}{3}}{\sqrt{1-\left(\frac{1}{3}\right)^2}}=\frac{\sqrt{2}}{4}$$
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 $a / \sqrt{b^2-a^2}$.