1

How do i simplify $\arccos(x)−\arcsin(x)$ for $x$ in $(−1,1)$

i got somewhere that...

$\sin(x)= \cos(\frac{\pi}{2}-x)$ so $\arccos(\sin(x))+x=\frac{\pi}{2}$

substituting that $\sin(x)=t \rightarrow \arcsin(t)=x$

$\arccos(t)+\arcsin(t)=\frac{\pi}{2}$

so working backwards $\arccos(t)-\frac{\pi}{2} = -(\arcsin(t))$

hence $\arccos(x)−\arcsin(x) = \arccos(t) + \arccos(t)-\frac{\pi}{2} = 2\arccos(t)-\frac{\pi}{2}$

Which hasn't really simplified anything

Ivan
  • 919
  • 4
  • 13
  • 27
  • 1
    Well, it's a little bit simpler, as it contains just one transcendental function, not two. There isn't much more you can do. By the way, you have mixed up the variables, ending up with a function of $x$ equalinga function of $t$. – Harald Hanche-Olsen Sep 03 '14 at 05:11
  • Conceivably the problem-setter made a slip, and intended $+$ and not $-$. – André Nicolas Sep 03 '14 at 05:14
  • You're basically asking for the difference between the two acute angles of a right triangle, which can be anything in between $0$ and $90$ degrees. – Lucian Sep 03 '14 at 05:15
  • @user156684, See also, http://math.stackexchange.com/questions/672575/proof-for-the-formula-of-sum-of-arcsine-functions-arcsin-x-arcsin-y setting $x=y$ – lab bhattacharjee Sep 03 '14 at 06:16

3 Answers3

2

$\arccos(x)$ is the angle $\theta$ with $\cos(\theta) = x$ and $0 \le \theta \le \pi$, while $\arcsin(x)$ is the angle $\phi$ with $\sin(\phi) = x$ and $-\pi/2 \le \phi \le \pi/2$. Now $\sin(\theta) = \cos(\pi/2 - \theta)$ and $-\pi/2 \le \pi/2 - \theta \le \pi/2$, so $\pi/2 - \theta = \phi$. Thus $$\arccos(x) - \arcsin(x) = \theta - \phi = 2\theta - \pi/2 = 2 \arccos(x) - \pi/2$$ or if you prefer $$ \pi/2 - 2 \phi = \pi/2 - 2 \arcsin(x)$$

Robert Israel
  • 448,999
  • Nice solution! It is probably better to write $\arccos(x) - \arcsin(x) = (3/2)\pi-2\theta$ so that $x=-1 \implies \theta=0$ and $x=+1 \implies \theta=\pi$. Thus $dx/d\theta>0$. – mike Sep 03 '14 at 05:20
  • So all this is trying to say is that graphically the angle of cos and the angle of sin add up to pi/2. And this question is trying to see if you understand that? – Ivan Sep 03 '14 at 05:40
2

This isn't a different conclusion from yours, but it is a different route. $\arccos(x)+\arcsin(x)$ is a constant $\frac{\pi}{2}$, which can be seen geometrically or with calculus. So then you have $$\arccos(x)-\left(\frac{\pi}{2}-\arccos(x)\right)=2\arccos(x)-\frac{\pi}{2}$$

2'5 9'2
  • 54,717
1

Setting $$ u(x)=\arccos x-\arcsin x \quad\forall x \in (-1,1), $$ we have $$ u'(x)=-\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}=-\frac{2}{\sqrt{1-x^2}}\quad\forall x \in (-1,1). $$ It follows that $$ \arccos x-\arcsin x=u(0)+\int_0^xu'(t)\,dt=\frac\pi2-2\int_0^x\frac{1}{\sqrt{1-t^2}}\,dt, $$ i.e. for every $x\in (-1,1)$ we have: $$ \arccos x-\arcsin x=\frac\pi2-2\arcsin x=\frac\pi2+2(\arccos x-\frac\pi2)=2\arccos x-\frac\pi2. $$

HorizonsMaths
  • 16,526