I present you my attempt to demonstrate that $4=0$ :

1) Beginning with the well-known identity $\cos ^{ 2 }{ x } =1-\sin ^{ 2 }{ x }$

2) $\cos { x } ={ (1-\sin ^{ 2 }{ x } ) }^{ 1/2 }$

3) $1+\cos { x } =1+{ (1-\sin ^{ 2 }{ x } ) }^{ 1/2 }$

4) ${ (1+\cos { x } ) }^{ 2 }={ (1+{ (1-\sin ^{ 2 }{ x } ) }^{ 1/2 }) }^{ 2 }$

5) Evaluate $x=\pi$ : ${ (1+\cos { \pi } ) }^{ 2 }={ (1+{ (1-\sin ^{ 2 }{ \pi } ) }^{ 1/2 }) }^{ 2 }$

6) ${ (1-1) }^{ 2 }={ (1+{ (1-0) }^{ 1/2 }) }^{ 2 }$

7) $0=4$

In which step is the first error committed?

Step 5
Step 1
Step 4
Step 7
Step 3
Step 2
Step 6

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

The formal definition of absolute value is the square root of the square:

$\sqrt{x^2}=|x|$

This should have been applied in step 2.