Spotting the Error: 0=4 in Math

[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.

 

[Infinitum]

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

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)|

 

[Mentallic]

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

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.