Let #A=18^@#, hence #5A=90^@# and
#2A=90^@-3A# and #sin2A=sin(90^@-3A)=cos3A#
or #2sinAcosA=4cos^3A-3cosA#
i.e. #cosA(2sinA-4cos^2A+3)=0# and as #cosA=cos18^@!=0#
we have #2sinA-4cos^2A+3=0#
or #2sinA-4(1-sin^2A)+3=0#
or #4sin^2A+2sinA-1=0# and hence
#sinA=(-2+-sqrt(2^2-4*4*(-1)))/8# or #(-2+-2sqrt5)/8# i.e. #(sqrt5-1)/4#
(as #m/_A=18^@#, #sinA# and #cosA# cannot be negative)
and #cosA=sqrt(1-(sqrt5-1)^2/16)=1/4sqrt(16-(5+1-2sqrt5)#
= #sqrt(10+2sqrt5)/4#
And #cos48^@=cos(30^@+18^@)#
= #cos30^@cos18^@-sin30^@sin18^@#
= #sqrt3/2xxsqrt(10+2sqrt5)/4-1/2xx(sqrt5-1)/4#
= #1/8(sqrt(30+6sqrt5)+1-sqrt5)#