# Proof 0=4

1. May 12, 2012

### powerof

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:

$sin^2 \theta + cos^2 \theta = 1 \Rightarrow cos \theta = \sqrt{1-sin^2 \theta}$​

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

$1 + cos \theta = 1 + \sqrt{1-sin^2 \theta}$

$(1 + cos \theta)^2 = (1 + \sqrt{1-sin^2 \theta})^2$​

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

Given that

$cos \pi=-1$

$sin \pi=0$​

it follows:

$(1-1)^2=(1+ \sqrt{1-0^2})^2$

$0=2^2 \Rightarrow 0=4$​

So, what's the problem here?

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

Have a nice day.

2. May 12, 2012

### Infinitum

I have a similar question in my book too, very fundamental error right in the first step.

For all real numbers x,

Therefore,

$\sqrt{cos^{2}(\theta)} = |cos(\theta)|$

Last edited: May 12, 2012
3. May 12, 2012

### Mentallic

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

Remember that the general rule is $$\sin^2\theta+\cos^2\theta=1$$ and so if we re-arrange to solve for $\cos\theta$ we would get $$\cos^2\theta=1-\sin^2\theta$$ 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, $$\cos\theta=\pm \sqrt{1-\sin^2\theta}$$ where we take the positive value when $\cos\theta >0$ and the negative when $\cos\theta <0$.