1. ## Inverse Trigonometry

Hello all,
just after a bit of help with this question, do i show this by taking the derivative? i'm not sure..

Show that
$\displaystyle \tan^{-1}x=\sin^{-1}\frac{x}{\sqrt {1+x^2}}$ for all $\displaystyle x \in (-\infty, \infty)$

2. $\displaystyle \tan v = \frac{\sin v}{ \cos v}=\frac{x}{1} \implies \arctan \frac{x}{1} = \arcsin \frac{x}{\sqrt{1+x^2}}$

As you can see above, $\displaystyle \sin v = x$ and $\displaystyle \cos v = 1$. The hypotenuse then becomes $\displaystyle \sqrt{1+x^2}$ (Pythagoras' theorem).

3. Hello, Robb!

No derivatives needed . . . It's an Identity problem.

Show that: .$\displaystyle \tan^{\text{-}1}x \;=\;\sin^{\text{-}1}\!\left(\frac{x}{\sqrt {1+x^2}}\right)$ . for all $\displaystyle x \in (\text{-}\infty, \infty)$
$\displaystyle \text{Let: }\:\theta \:=\:\tan^{\text{-}1}x$ .[1]

Then we have: .$\displaystyle \tan\theta \:=\:\frac{x}{1} \:=\:\frac{opp}{adj}$

$\displaystyle \theta$ is in a right triangle with: $\displaystyle opp = x,\;adj = 1$

Using Pythagorus, we have: .$\displaystyle hyp = \sqrt{1+x^2}$

Hence: .$\displaystyle \sin\theta \:=\:\frac{opp}{hyp} \:=\:\frac{x}{\sqrt{1+x^2}} \quad\Rightarrow\quad \theta \:=\:\sin^{\text{-}1}\!\left(\frac{x}{\sqrt{1+x^2}}\right)$ .[2]

Equate [1] and [2]: . $\displaystyle \tan^{\text{-}1}x \;=\;\sin^{\text{-}1}\!\left(\frac{x}{\sqrt{1+x^2}}\right)$