I'm doing IVT proofs and I'm trying to prove there's f(x)=0 for cosx=arcosx. I thought for sure it would equal f(x)=cos(cosx)-x. I get this by first by cosx=arcosx, cos(cosx)=x, cos(cosx)-x=0. I'm not sue though.... Is cosx=arcosx really just cosx-arcosx=0?? This is my first post!! Thanks.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Setting cosx=arccosx equal to zero.
- Thread starter Mandrew98
- Start date
You're right but you need to be careful about your domain. For \(\displaystyle -1\le x\le 1\)
\(\displaystyle \cos(x)=\arccos(x) \iff f(x)=\cos(\cos(x))-x=0\)
Edit: Turns out my hint was no good, its far easier than I thought. \(\displaystyle 0<cos(cos(x))<1\) in \(\displaystyle [0,1]\) so what is the sign of \(\displaystyle f(1)\)?
\(\displaystyle \cos(x)=\arccos(x) \iff f(x)=\cos(\cos(x))-x=0\)
Edit: Turns out my hint was no good, its far easier than I thought. \(\displaystyle 0<cos(cos(x))<1\) in \(\displaystyle [0,1]\) so what is the sign of \(\displaystyle f(1)\)?
Last edited:
Thanks. I was going to say f(0)=.540 and f(1)=-.142 so by IVT f(0)>0>f(1) so c € E [0,1]where f(c)=0. I think your way is better cause you don't need a calculator!! Thanks again, I'm very excited to use this forum!!
Yes, this is fine