# Thread: Proof that two inverse trig functions are equal for all X

1. ## Proof that two inverse trig functions are equal for all X

I seem to be having some issues.
Prove that:

$\displaystyle 2arcsin(x) = arccos(1-2x^2) \forall x>0$

If the above is true, then

$\displaystyle 2arcsin(x) - arccos(1-2x^2) = f(x) = 0 \forall x>0$

$\displaystyle f(1) = 0$

but

$\displaystyle f(2) \not= 0$
by WolframAlpha/Mathematica
http://www.wolframalpha.com/input/?i=2+arcsin%282%29++-+arccos%281-2%282%29^2%29

Can anyone figure out what I am doing incorrectly?

2. Originally Posted by skyd171
I seem to be having some issues.
Prove that:

$\displaystyle 2arcsin(x) = arccos(1-2x^2) \forall x>0$

If the above is true, then

$\displaystyle 2arcsin(x) - arccos(1-2x^2) = f(x) = 0 \forall x>0$

$\displaystyle f(1) = 0$

but

$\displaystyle f(2) \not= 0$
by WolframAlpha/Mathematica
http://www.wolframalpha.com/input/?i=2+arcsin%282%29++-+arccos%281-2%282%29^2%29

Can anyone figure out what I am doing incorrectly?
Obviously the domain of f(x) is $\displaystyle -1 \leq x \leq 1$.

3. Hello, skyd171!

$\displaystyle \text{Prove that: }\;2\arcsin(x) \:=\: \arccos(1-2x^2),\;\;\forall x>0$

$\displaystyle \text{Let: }\:\alpha \,=\, \arcsin(x)$

The left side is: .$\displaystyle 2\alpha$

Take the cosine: .$\displaystyle \cos(2\alpha) \:=\:2\cos^2\alpha-1$ .[1]

$\displaystyle \text{Since }\,\sin\alpha \:=\:x \;=\; \dfrac{x}{1} \:=\:\dfrac{opp}{hyp}$

. . $\displaystyle \,\alpha$ is is a right triangle with: $\displaystyle opp = x,\;hyp = 1$

$\displaystyle \text{Pythagorus gives us: }\:adj \,=\,\sqrt{1-x^2}$

. . $\displaystyle \text{Hence: }\:\cos\alpha \:=\:\sqrt{1-x^2}$

Substitute into [1]:

. . $\displaystyle \cos(2\alpha) \;=\;2(\sqrt{1-x^2})^2 - 1 \;=\;2(1-x^2)-1 \;=\;2-2x^2 - 1$

. . $\displaystyle \cos(2\alpha) \;=\;1 - 2x^2$

$\displaystyle \text{Then: }\:2\alpha \;=\;\arccos(1-2x^2)$

$\displaystyle \text{Therefore: }\:2\arcsin(x) \;=\;\arccos(1-2x^2)$

4. Thanks all.