Question

Calculati:
$\frac{1-cos \alpha + isin \alpha }{1+cos \alpha -isin \alpha }$

$\frac{1+cos \alpha -isin \alpha }{1-cos \alpha +isin \alpha }$

$\frac{1+cos \alpha +isin \alpha }{1+cos \alpha -isin \alpha }$

Answer 1

stim          1 -cosα = cos0 - cosα = -2sin(0 - α) /2 ·sin(0 +α ) /2 = 2sin²α/2
                 1 +cosα =cos0 + cosα = 2cos²α/2 
                   sinα = sin(2α/2) = 2sinα/2·cosα/2 
ex este pentru  α ∈( 0 ; π/2) 
a.   [ 2sin²α/2    + 2isinα/2·cosα/2]  / [ 2cos²α/2 - 2isinα/2·cosα/2]= 
         = 2sinα/2 · [ sinα/2  + icosα/2] / 2cosα/2 ·[ cosα/2  - isinα/2] =
                     sinα/2 = cosα/2   si cosα/2 = sinα/2 in cadranul I
           = tgα/2 · [cosα/2 + isinα/2] / [ cos( -α/2) + isin(-α/2] 
                  = tgα/2· [ cos ( α/2 - (-α/2) )   + i sin ( α/2 - (-α/2))]
                   = tgα/2 ·[ cosα + isinα]
b. [ 2cos²α/2 - 2isinα/2·cosα/2] / [ 2sin²α/2 +2isinα/2·cosα/2]=
     = 2cosα/2 ·[ cosα/2 - isinα/2] / 2sinα/2·[sinα/2  + icosα/2]=
      = ctgα/2 · [  cos(-α/2)  + isin(-α/2) ] / [cosα/2 + isinα/2]
       = ctgα/2 · [ cos( -α/2 -α/2)  + isin( -α/2 - α/2) ]
       = ctgα/2 · [ cos( -α) + isin(-α)]
       =  ctgα/2 · [ cosα - isinα]
cosα = para  ; sinα=impara     ; cos(-α) = cosα  ; sin(-α) =- sinα
c.   [2cos²α/2 +2isinα/2 ·cosα/2]  / [ 2cos²α/2 -2isinα/2·cosα/2]=
      = 2cosα/2 ·[ cosα/2 + isinα/2 ] / 2cosα/2·[cosα/2 -isinα/2]=
        = 2cosα/2  : 2cosα/2  · [ cosα/2 +isinα/2 ] / [cosα/2 - i sinα/2]
        = [ cosα/2 + i sinα/2 ] / [ cos ( - α/2)  + isin( - α/2) ]
        =  cos[α /2  - ( -α/2) ] + i sin[ α/2 - ( - α/2)]
         =  cosα + isinα