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It's been awhile since I've used trig and I feel stupid asking this question lol but here goes:

Given: $z = \tan(\arcsin(x))$

Question: How do I write something like that in terms of $x$?

Thanks! And sorry for my dumb question.

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up vote 2 down vote accepted

$$z = \tan[\arcsin(x)]$$

$$\arctan(z) = \arctan[\tan(\arcsin(x)] = \arcsin(x)$$

$$\sin[\arctan(z)] = \sin[\arcsin(x)] = x$$

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So you basically just do the inverse of each side until x is by itself? – ModdedLife Apr 21 '13 at 21:29
I'm assuming you want x as a function of $z$? To isolate $x$? – amWhy Apr 21 '13 at 21:34
Yes, that is correct. – ModdedLife Apr 21 '13 at 21:35
Thenwh at I've written above will indeed give you x as a function of $z$: $$x = \sin[\arctan(z)]$$ – amWhy Apr 21 '13 at 21:39
Ok thanks. So if the question was worded something like: "Find the value of z = tan(arcsin(x)) in terms of x." does this mean the same thing that you have shown here? Or would that be more like what Andre showed as the answer? – ModdedLife Apr 21 '13 at 21:42

Draw a triangle $ABC$, right-angled at $C$. Let us suppose that $\arcsin x=\angle A$.

Decide that the hypotenuse $AB$ is of length $1$ (it doesn't matter). Write $1$ next to the hypotenuse.

Then the side $BC$ opposite $\angle A$ has length $x$. The remaining side $AC$, by the Pythagorean Theorem, is $\sqrt{1-x^2}$. Write the appropriate lengths on the diagram.

Finally, read off $\tan A$ from the picture. It is $\frac{x}{\sqrt{1-x^2}}$.

This is not quite all there is to it. We need to check whether the result makes sense for all $x$. The only possible issues that can arise for angles not between $0$ and $\frac{\pi}{2}$ are issues of sign. Since $\arctan x$ is conventionally always between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, and on this interval $\sin$ has the same sign as $\tan$, there is no problem.

If you want something like $\csc(\arctan x)$, do something similar. We could let the hypotenuse be $1$. But it may be easier to let the leg adjacent to $\angle A$ be $1$, and the leg opposite $\angle A$ be $x$. Then the hypotenuse is $\sqrt{1+x^2}$, and now you can read off the cosecant from the picture.

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Ok thank you. So would that be "the value of z in terms of x?" – ModdedLife Apr 21 '13 at 21:33
OK this made perfect sense now. Thanks for your help! – ModdedLife Apr 21 '13 at 21:53
The expression you write already gives the value of $z$ in terms of $x$, but in a complicated way. There are quite a few situations in which we want an algebraic simplification of your expression. This happens most frequently when we have made a trigonometric substitution to evaluate an integral. But it is very possible that you might be asked for this kind of simplification in a "precalculus" course. – André Nicolas Apr 21 '13 at 22:00

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