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$$\arccos\left(\frac{1-x^2}{1+x^2}\right) = 2\arctan{x}$$ for $x \geq 0$.

I'm not even sure what kind of math to try? :(

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up vote 4 down vote accepted

Put $t=\arctan x$. Then $x=\tan t$. You have to prove that $2t=\arccos(1-x^2)/(1+x^2)$. This is more-or-less the same as $(1-x^2)/(1+x^2)=\cos2t$. If you put $x=\tan t$ into $(1-x^2)/(1+x^2)$ what do you get?

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Setting $x=\tan a$ your equality

$$\arccos \dfrac{1-x^{2}}{1+x^{2}}=2\arctan x\qquad (1)$$


$$\cos 2a=\dfrac{1-\tan ^{2}a}{1+\tan ^{2}a}.\qquad (2)$$

To prove (2) which is listed in the wikipedia we can take the duplication formula

$$\cos 2a=\cos ^{2}a-\sin ^{2}a$$

Dividing by $\cos ^{2}a+\sin ^{2}a=1$, we establish

$$\cos 2a=\dfrac{\cos ^{2}a-\sin ^{2}a}{\cos ^{2}a+\sin ^{2}a}=\dfrac{1-\tan ^{2}a}{1+\tan ^{2}a}.$$

Edit: One must note that $\arctan x$ is an odd function, while $\arccos \frac{1-x^{2}}{1+x^{2}}$ is an even one. Only for $x\geq 0$ are both sides of $(1)$ equal. Identity $(2)$ is valid for both positive and negative values of $a$.

alt text

Blue: $y=\arccos((1-x^2)/(1+x^2))$ for $x\ge 0$; Red: $y=2\arctan x$

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Thats what Robin answered? – anonymous Sep 13 '10 at 12:51
@Chandru1: similar, yes. I based my deduction in a general method I learned that all direct trigonometric function of the angle $2a$ may be expressed as rational function of $\tan a$. – Américo Tavares Sep 13 '10 at 12:59
@Chandru: The book is "Compêndio de Trigonometria" by J. Jorge Calado, Lisbon, Empresa Literária Fluminence, 1967. – Américo Tavares Sep 13 '10 at 13:03

Differentiate the difference $$\arccos\left(\frac{1-x^2}{1+x^2}\right) - 2\arctan{x},$$ simplify the resulting expression and find $0$ when $x>0$, concluding that the difference is constant. Now evaluate at $1$, to see that the difference is in fact constantly zero.

N.B.: This is of course quite unenlightenling, but you always know that if two things are equal, their difference is constant so you never lose much by trying to show that it is constant and zero. Specially if you have a computer algebra system at hand that can do the derivatives and simplifications for you!

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