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Math Help - Trig Identities

  1. #1
    Member classicstrings's Avatar
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    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)
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  2. #2
    Junior Member
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    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)
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