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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?

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migrated from mathematica.stackexchange.com Nov 2 '13 at 13:26

This question came from our site for users of Wolfram Mathematica.

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    $\begingroup$ 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? $\endgroup$ – halirutan Nov 2 '13 at 12:28
  • $\begingroup$ @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. $\endgroup$ – Michael E2 Nov 2 '13 at 12:32
  • $\begingroup$ possible duplicate of How to derive compositions of trigonometric and inverse trigonometric functions? $\endgroup$ – Git Gud Nov 2 '13 at 13:28
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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

enter image description here

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Here is how you would do it if you didn't have Wikipedia.

$$\tan\left(\arcsin\left(\frac{1}{3}\right)\right) =\frac{\sin\left(\arcsin\left(\frac{1}{3}\right)\right)}{\cos\left(\arcsin\left(\frac{1}{3}\right)\right)} =\frac{\sin\left(\arcsin\left(\frac{1}{3}\right)\right)}{\sqrt{1-\sin^2\left(\arcsin\left(\frac{1}{3}\right)\right)}} =\frac{\frac{1}{3}}{\sqrt{1-\left(\frac{1}{3}\right)^2}}=\frac{\sqrt{2}}{4}$$

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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 $a / \sqrt{b^2-a^2}$.

Mathematica graphics

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  • $\begingroup$ Nicely done, Michael! +1 $\endgroup$ – Namaste Nov 2 '13 at 13:43

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