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To prove:

$$\begin{align} \sin({\arccos{x}})&=\sqrt{1-x^2}\\ \cos{\arcsin{x}}&=\sqrt{1-x^2}\\ \sin{\arctan{x}}&=\frac{x}{\sqrt{1+x^2}}\\ \cos{\arctan{x}}&=\frac{1}{\sqrt{1+x^2}}\\ \tan{\arcsin{x}}&=\frac{x}{\sqrt{1-x^2}}\\ \tan{\arccos{x}}&=\frac{\sqrt{1-x^2}}{x}\\ \cot{\arcsin{x}}&=\frac{\sqrt{1-x^2}}{x}\\ \cot{\arccos{x}}&=\frac{x}{\sqrt{1-x^2}} \end{align}$$

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Draw right angled Triangles – lab bhattacharjee Jun 21 '13 at 18:55
I got speak my mind about the above tip: to me it's worthless. Drawings might be helpful to help one think, but I don't count them as proof and neither do the people who mark exams. – Git Gud Jun 21 '13 at 19:25
@GitGud: Why shouldn't drawings count as proof? As long as you remember that the proof only works for angles smaller than $90^\circ$, they're perfectly good arguments. – Javier Jun 21 '13 at 21:09
@JavierBadia I draw the line for what counts as a proof way before drawings are reached. Nothing but a formal proof is an actual proof so any given person will be satified with what he/she will be satisfied with. Drawigns mean nothing to me and I'm sure you've encountered multiple examples where drawings are misleading. Furthermore, they often don't count as proof for people marking exams and anyone who asks this question is at a level that he should be worried about how to deal with the details. Finally, personally, I hate visual mathematics. – Git Gud Jun 21 '13 at 21:14
up vote 7 down vote accepted

I'll be ignoring domains and possible roots of negative numbers. (If you let $\mbox{$x\in \textbf{]}0,\pi/2[$}$ everything works fine).

Given $f\circ g$, the trick is too relate $f$ with $g^{-1}$.

I did some. You should be able to handle the remaining ones.

$\bullet \sin (\arccos (x))=\sqrt {1-(\cos (\arccos (x) ))^2}=\sqrt {1-x^2}$

$\bullet \sin (\arctan (x))=\dfrac{\tan (\arctan (x))}{\sqrt {1+(\tan (\arctan (x)))^2}}=\dfrac{x}{\sqrt {1+x^2}}$

For this one I used $$\begin{align}(\cos(x))^2+(\sin (x))^2=1 &\implies 1+(\tan(x))^2=(\sec(x))^2\\ &\implies 1+(\tan (x))^2=\dfrac{1}{1-(\sin (x))^2}\\ &\implies 1-(\sin(x))^2=\dfrac{1}{1+(\tan(x))^2}\\ &\implies \sin (x)=\dfrac{\tan(x)}{\sqrt{1+(\tan(x))^2}}\end{align}$$

$\bullet \tan (\arcsin (x))=\dfrac{\sin (\arcsin (x))}{\sqrt{1-(\sin(\arcsin (x)))^2}}=\dfrac{x}{\sqrt{1-x^2}}$

For this one I used $$\begin{align}(\cos(x))^2+(\sin (x))^2=1 &\implies 1+(\tan(x))^2=(\sec(x))^2\\ &\implies 1+(\tan (x))^2=\dfrac{1}{1-(\sin (x))^2}\\ &\implies \tan(x)=\dfrac{\sin(x)}{\sqrt{1-(\sin(x))^2}}\end{align}$$

$\bullet \cot (\arcsin(x))=\dfrac{\sqrt{1-(\sin (\arcsin(x)))^2}}{\sin (\arcsin(x))}=\dfrac{\sqrt{1-x^2}}{x}$

For this one I used $$\begin{align}(\cos(x))^2+(\sin (x))^2=1 &\implies (\cot (x))^2+1=(\csc(x))^2\\ &\implies (\cot (x))^2=\dfrac{1-(\sin(x))^2}{(\sin(x))^2}\\ &\implies \cot(x)=\dfrac{\sqrt{1-(\sin(x))^2}}{\sin (x)}\end{align}$$

$\bullet \cot (\arccos(x))=\dfrac{\cos (\arccos(x))}{\sqrt{1-(\cos (\arccos(x)))^2}}=\dfrac{x}{\sqrt{1-x^2}}$.

For this one I used $$\begin{align}(\cos(x))^2+(\sin (x))^2=1 &\implies (\cot (x))^2+1=(\csc(x))^2\\ &\implies (\cot (x))^2=\dfrac{1-(\sin(x))^2}{(\sin(x))^2}\\ &\implies (\cot(x))^2=\dfrac{(\cos(x))^2}{1-(\cos(x))^2}\\ &\implies \cot(x)=\dfrac{\cos(x)}{\sqrt{1-(\cos(x))^2}}\end{align}$$

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Could you prove $\cot(x)$ in terms of $\cos(x)$ only ? Is it possible to get $\cot(x)$ in terms of $x$ only ? Thanks. – Nicolas Lykke Iversen Nov 24 '13 at 10:55
@NicolasLykkeIversen Done. Yes, it is always possible on account of the very definition of $\cot\color{grey}{=\dfrac{\cos}{\sin}}$. The definition itself gives a much quicker derivation of $\cot(x)=\dfrac{\cos(x)}{\sqrt{1-(\cos(x))^2}}$ than what I did on the answer. I chose to do it the way I did on the answer to keep a certain homogeneity between the derivation of the identities. – Git Gud Nov 24 '13 at 11:10

Here is astart

$$ \sin({\arccos{x}})= \sqrt{1-\cos(\arccos(x))^2}=\sqrt{1-x^2}, $$

since $\cos(\arccos(x))=x.$ You need to use some trig. identities.

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I agree with the comment about right angled triangles. For example, by definition, $\arccos(x)$ is the angle $\theta$ in a right triangle with hypotenuse $1$ occurring adjacent to a leg of length $x$, and by the Pythagorean theorem and the definition of sine, the sine of such an angle is $\sqrt{1-x^2}/1$ (draw the picture). The others can be shown similarly.

I think this is perfectly rigorous (and can be extended by symmetry or differentiation arguments to the whole domain of the arccosine function) -- it's just a matter of what you take as the starting definition of the trig functions.

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@coutinghaus The problem is that geometric definitions are untreatable. You gotta go from the plane, to the plane with the axis, define angle, define triangle, etc..., then extend the functions by periodicity. I don't really know how to do it, maybe I'm wrong, but it seems very problematic to me. – Git Gud Jun 21 '13 at 19:38
I believe that the proof by user29743 is the best one. Using this method one can derive several identities involving trig and inverse trig functions which may be useful for integration for example. – Idris Mar 26 '15 at 23:42

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