Thread: Trig Identities

Given that z = cos@ + isin@, where -pie < @ < pie, show that

1 + z = 2cos^2(@/2) + 2isin(@/2)cos(@/2).

Hence show that mod( 1 + z) = 2cos(@/2)

1=cos0+isin0
1+z=(cos0+isin0)+(cos@+isin@)
=(cos@+cos0)+i(sin@+sin0)
=2cos((@+0)/2)cos((@-0)/2)+i(2sin((@+0)/2)cos((@-0)/2)
=2cos^2(@/2)+2isin(@/2)cos(@/2)

mod(1+z)=sqrt((2cos^2(@/2)^2+(2sin(@/2)cos(@/2))^2)
mod(1+z)=sqrt{4cos^(@/2)[cos^2(@/2)+sin^2(@/2)]}
=2cos(@/2)sqrt(cos^2(@/2)+sin^2(@/2))
=2cos(@/2)(1)
=2cos(@/2)