To calculate #cos(a/2)# and #sin(a/2)#, apply the formula
#cos^2(a/2)=(1+cosa)/2# and #sin^2(a/2)=(1-cosa)/2#.
First, you have to know the value #cosa#.
#cos^2a=1-sin^2a=1-0.75^2=0.4375=7/16#
#cosa=+-sqrt(7)/4=+-0.6614#
[Case1] If #cosa=sqrt(7)/4#, then
#cos^2(a/2)=(4+sqrt(7))/8# and #sin^2(a/2)=(4-sqrt(7))/8#
#cos(a/2)=+-sqrt((4+sqrt(7))/8)=+-sqrt(8+2sqrt(7))/4=+-(sqrt(7)+1)/4#
#sin(a/2)=+-sqrt((4-sqrt(7))/8)=+-sqrt(8-2sqrt(7))/4=+-(sqrt(7)-1)/4#
Note that #cos(a/2)# and #sin(a/2)# must have the same sign because #sina=2sin(a/2)cos(a/2)# is positive(0.75).
Therefore, #(cos(a/2),sin(a/2))=+-((sqrt(7)+1)/4, (sqrt(7)-1)/4)=+-(0.9114,0.4114)# is the answer in [Case 1].
[Case2] If #cosa=-sqrt(7)/4#, then
#cos^2(a/2)=(4-sqrt(7))/8# and #sin^2(a/2)=(4+sqrt(7))/8#.
In this case, #cos^2(a/2)# and #sin^2(a/2)# exchanged each other compared to [Case1].
The value for #cos(a/2)# and #sin(a/2)# are also exchanged.
#(cos(a/2),sin(a/2))=+-((sqrt(7)-1)/4, (sqrt(7)+1)/4)=+-(0.4114,0.9114)#
