Let #A=36^@#, hence #5A=180^@# and
#2A=180^@-3A# and #sin2A=sin(180^@-3A)=sin3A#
or #2sinAcosA=3sinA-4sin^3A# and as #sinA!=0#, dividing by it we get
#2cosA=3-4sin^2A# and #2cosA=3-4(1-cos^2A)#
i.e. #4cos^2A-2cosA-1=0#
and using quadratic formula #cosA=(2+-sqrt(2^2-4xx4xx(-1)))/8#
= #(2+-sqrt20)/8=(sqrt5+1)/4#
As #cos36^@=sin54^@=(sqrt5+1)/4#
and #cos54^@=sqrt(1-(sqrt5+1)^2/16)=1/4sqrt(16-(5+1+2sqrt5)#
= #sqrt(10-2sqrt5)/4#
#sin84^@=sin(54^@+30^@)#
= #sin54^@cos30^@+cos54^@sin30^@#
= #(sqrt5+1)/4xxsqrt3/2+sqrt(10-2sqrt5)/4xx1/2#
= #1/8(sqrt15+sqrt3+sqrt(10-2sqrt5))#