inverse trig functions

**xxju1anxx** · October 7th 2009, 12:38 PM

how can I simplify this: tan(arcsin(x))

**Soroban** · October 7th 2009, 12:58 PM

Hello, xxju1anxx!

Simplify: . $\tan\bigg[\arcsin(x)\bigg]$

An inverse trig function is an angle.

Let $\theta \,=\,\arcsin(x) \quad\Rightarrow\quad\sin\theta \,=\,x$

We have: . $\sin\theta \:=\:\frac{x}{1} \:=\:\frac{opp}{hyp}$

$\theta$ is in a right triangle with: . $opp = x,\;hyp = 1$
. . Using Pythagorus, we find that: . $adj = \sqrt{1-x^2}$
Hence: . $\tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{x}{\sqrt{1-x^2}}$

Therefore: . $\tan\bigg[\arcsin(x)\bigg] \;=\;\tan(\theta) \;=\;\frac{x}{\sqrt{1-x^2}}$

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