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I have this function, $$\sin\left[{\arctan\left({\frac{x}{\sqrt{1-x^2}}}\right)}\right]$$

I drew a right angled triangle putting $x$ on the opposite side and the square root on the adjacent which makes the hypotheses being $1$, and since that $\sin = \frac{\text{opposite }(x)}{\text{hypotenuse (1)}}$ the result is equal to $x$, now my question is the result doesn't respect the domain as the original function am I supposed to add one and am I supposed to reverse the denominator and nominator in the case of an arc function? because when I do that I get $\sqrt{1-x^2}$ which isn't correct

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The identity function does in fact respect the domain of your function, since you need $\sqrt {1-x^2}>0$, you get that $-1<x<1$ is your domain, which fits the domain/range of $f(x)=x$ in this case (since $\sin$ outputs $[-1,1]$

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  • $\begingroup$ Oh right, but what about the inverse part? do I just ignore it? like if the function was sin of tan it would have been same steps $\endgroup$
    – method
    Commented Dec 15, 2014 at 3:30
  • $\begingroup$ Inverse part? Arctan can take any input, but the $\sqrt$ in your denominator of the input of arctan is the only real restriction. Sin of tan wouldn't be the same steps, because you only draw the triangle if the inner function is an inverse trig function $\endgroup$
    – Alan
    Commented Dec 15, 2014 at 3:32
  • $\begingroup$ Oh so if it was sin of tan there is no evaluating could be done? also one last question, if I was asked to differentiate such funcion am I supposed to evaluate first? $\endgroup$
    – method
    Commented Dec 15, 2014 at 3:34
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    $\begingroup$ You're not 'evaluating' it, evaluating would be putting in a specific value of x and getting an output. This is more just 'simplifying', i.e. rewriting the function in a simpler form. And you should always simplify before doing calculus if possible. And yes, there's no simplifying sin of tan to my knowledge. $\endgroup$
    – Alan
    Commented Dec 15, 2014 at 3:41
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Let $\arctan\dfrac x{\sqrt{1-x^2}}=u\implies\tan u=\dfrac x{\sqrt{1-x^2}},$

and $ -\dfrac\pi2\le u\le\dfrac\pi2\implies\cos u\ge0$

$\implies\cos u=+\dfrac1{\sqrt{1+\tan^2u}}=\sqrt{1-x^2}$

and consequently, $\sin\left(\arctan\dfrac x{\sqrt{1-x^2}}\right)=\sin u=\cos u\cdot\tan u=?$

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