1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proof 0=4

  1. May 12, 2012 #1
    I found this in my math book. It gives this supposed proof that 0=4 and asks where the error is. Note that this isn't homework. I found it at the end of the unit among other bonus problems to supposedly test ingenuity. Here it is:

    [itex]sin^2 \theta + cos^2 \theta = 1 \Rightarrow cos \theta = \sqrt{1-sin^2 \theta}[/itex]​

    Now we add 1 to both sides and then square them.

    [itex]1 + cos \theta = 1 + \sqrt{1-sin^2 \theta}[/itex]

    [itex](1 + cos \theta)^2 = (1 + \sqrt{1-sin^2 \theta})^2[/itex]​

    Next we substitute theta with pi.
    [itex](1 + cos \pi)^2 = (1 + \sqrt{1-sin^2 \pi})^2[/itex]​

    Given that

    [itex]cos \pi=-1[/itex]

    [itex]sin \pi=0[/itex]​

    it follows:

    [itex](1-1)^2=(1+ \sqrt{1-0^2})^2[/itex]

    [itex]0=2^2 \Rightarrow 0=4[/itex]​

    So, what's the problem here?

    Thanks for taking time to read this and hopefully solve it.

    Have a nice day.
  2. jcsd
  3. May 12, 2012 #2
    I have a similar question in my book too, very fundamental error right in the first step.

    For all real numbers x,



    [itex]\sqrt{cos^{2}(\theta)} = |cos(\theta)|[/itex]
    Last edited: May 12, 2012
  4. May 12, 2012 #3


    User Avatar
    Homework Helper

    This rule when stated usually implies that [tex]\theta\in \left[-\frac{\pi}{2},\frac{\pi}{2}\right][/tex] but it works for whenever [itex]\cos\theta \geq 0[/itex]

    Remember that the general rule is [tex]\sin^2\theta+\cos^2\theta=1[/tex] and so if we re-arrange to solve for [itex]\cos\theta[/itex] we would get [tex]\cos^2\theta=1-\sin^2\theta[/tex] and at this point, if we're to take the square root of both sides, we need to keep in mind that there is also a negative square root value as well, mainly, [tex]\cos\theta=\pm \sqrt{1-\sin^2\theta}[/tex] where we take the positive value when [itex]\cos\theta >0[/itex] and the negative when [itex]\cos\theta <0[/itex].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook