Inverse Trig Identity

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July 10th 2012, 08:37 AM
Greymalkin

Inverse Trig Identity

The actual problem is: prove that $sin^-1x=tan^-1{x\over {\sqrt{1-x^2}}$

My problem is: how does $sin(tan^-1{x\over \sqrt{1-x^2}})$ equal x??
I understand how $sin(sin^-1x)=x$ but not the rational inverse, why is it so???
July 10th 2012, 04:36 PM
emakarov

Re: Inverse Trig Identity

The idea is that if $x = \sin\alpha$ , then $\sin^{-1}x=\alpha$ . We have $\frac{\sin\alpha}{\sqrt{1-\sin^2\alpha}}=\tan\alpha$ , so $\tan^{-1}\frac{x}{\sqrt{1-x^2}}$ is indeed $\alpha$ .

One has to take care of various subtleties such as $\sqrt{x^2}=|x|$ rather than $x$ and $\sin^{-1}(\sin\alpha)=\alpha$ only for $\alpha\in[-\pi/2,\pi/2]$ .
July 10th 2012, 06:32 PM
Soroban

Re: Inverse Trig Identity

Hello, Greymalkin!

Quote:

$\text{Prove: }\:\sin^{\text{-}1}\!x\:=\:\tan^{\text{-}1}\!\left(\frac{x}{\sqrt{1-x^2}}\right)$

$\text{Let }\,\theta \:=\:\sin^{\text{-}1}\!x \;\;[1]$

$\text{Then: }\,\sin\theta \:=\:x \:=\:\frac{x}{1} \:=\:\frac{opp}{hyp}$

$\text{We see that }\theta\text{ is in a right triangle with: }\:\begin{Bmatrix}opp \:=\:x \\ hyp \:=\:1 \end{Bmatrix}$
$\text{From Pythagorus: }\:adj \:=\:\sqrt{1-x^2}$

$\text{Then: }\:\tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{x}{\sqrt{1-x^2}}$

$\text{Hence: }\:\theta \:=\:\tan^{\text{-}1}\!\left(\frac{x}{\sqrt{1-x^2}}\right) \;\;[2]$

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

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