4=0

Geometry Level 3

I present you my attempt to demonstrate that 4=04=0:

1) Beginning with the well-known identity cos2x=1sin2x\cos ^{ 2 }{ x } =1-\sin ^{ 2 }{ x }

2) cosx=(1sin2x)1/2\cos { x } ={ (1-\sin ^{ 2 }{ x } ) }^{ 1/2 }

3) 1+cosx=1+(1sin2x)1/21+\cos { x } =1+{ (1-\sin ^{ 2 }{ x } ) }^{ 1/2 }

4) (1+cosx)2=(1+(1sin2x)1/2)2{ (1+\cos { x } ) }^{ 2 }={ (1+{ (1-\sin ^{ 2 }{ x } ) }^{ 1/2 }) }^{ 2 }

5) Evaluate x=πx=\pi : (1+cosπ)2=(1+(1sin2π)1/2)2{ (1+\cos { \pi } ) }^{ 2 }={ (1+{ (1-\sin ^{ 2 }{ \pi } ) }^{ 1/2 }) }^{ 2 }

6) (11)2=(1+(10)1/2)2{ (1-1) }^{ 2 }={ (1+{ (1-0) }^{ 1/2 }) }^{ 2 }

7) 0=40=4

In which step is the first error committed?

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