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I have this trig function:

$$\sin(\tan^{-1}x)$$

How does one simplify this? I don't really get my book's approach and why we're doing it.

So I guess we're supposed to start by doing this:

$$\tan^{-1}x = y$$

so... $$\tan y = x$$

We're supposed to find the sin of that angle y, but how do we do this? Well, I have this trig identity: $\cot^2x + 1 = \csc^2x$... but I'm not sure where this gets me.

Graphically, I think the answer is $x/\sqrt{1+x^2}$ because if the tan of an angle is x, then that means opposite an angle there's an x, and then the other side is 1, and so the sin of that angle is $x/\sqrt{1+x^2}$ Is that right?

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  • $\begingroup$ How is this different from the question that you posted a few minutes previous to this? $\endgroup$ – amd Feb 4 '18 at 7:17
  • $\begingroup$ it's the converse? $\endgroup$ – Jwan622 Feb 4 '18 at 7:43
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$$|\sin\arctan{x}|=\sqrt{1-\frac{1}{1+\tan^2\arctan{x}}}=\frac{|x|}{\sqrt{1+x^2}}$$ Thus, since $\sin$ and $\arctan$ they are odd functions we got the answer: $$\sin\arctan{x}=\frac{x}{\sqrt{1+x^2}}$$

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  • $\begingroup$ How did we get to the big square root sign? $\endgroup$ – Jwan622 Feb 4 '18 at 6:27
  • $\begingroup$ I took $|.|$, which gives the sing $+$. $\endgroup$ – Michael Rozenberg Feb 4 '18 at 6:29
  • $\begingroup$ becase ur looking at $\sin(x)$ and not $\sin^2(x)$ $\endgroup$ – Asim90 Feb 4 '18 at 6:29
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    $\begingroup$ I used $|\sin\alpha|=\sqrt{1-\cos^2\alpha}=\sqrt{1-\frac{1}{1+\tan^2\alpha}}.$ $\endgroup$ – Michael Rozenberg Feb 4 '18 at 6:31
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    $\begingroup$ Under $\sqrt{}\,$ he put $1-\cos^2(x)$ and expressed $\cos^2(x)$ in terms of $\tan(x)$ using identity $\tan^2(x)=\sec^2(x)-1$ $\endgroup$ – Asim90 Feb 4 '18 at 6:32

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