We are to evalute #sin33#.
#sin33=sin(18+15)#
#=sin18cos15+cos18sin15.....(1)#
Evaluation of #sin18 and cos18#
Let #A=18^@#
#=>5A=90^@#
#=>3A=90^@-2A#
#:.cos(3A)=cos(90^@-2A)#
#=>4cos^3A-3cosA=sin2A=2sinAcosA#
#=>4cos^2A-3=2sinA#
#=>4-4sin^2A-3=2sinA#
#=>4sin^2+2sinA-1=0#
#=>sinA=(-2+sqrt(2^2-4*4*
(-1)))/(2*4)#
#=(-2+sqrt20)/8=(sqrt5-1)/4#
#:.sin(18^@)=(sqrt5-1)/4#,
#cos18^@=sqrt(1-sin^2 18^@)#
#=sqrt(1-(sqrt5-1)^2/16)#
#=sqrt(16-5-1+2sqrt5)/4#
#=1/4sqrt(10+2sqrt5)#
Evaluation of #sin15 and cos15#
#sin15=sqrt(1/2(1-cos30))#
#=sqrt(1/2(1-sqrt3/2))#
#=sqrt(1/8(4-2sqrt3))#
#=sqrt(1/8(sqrt3-1)^2)#
#=(sqrt3-1)/(2sqrt2)#
#cos15=sqrt(1/2(1+cos30))#
#=sqrt(1/2(1+sqrt3/2))#
#=sqrt(1/8(4+2sqrt3))#
#=sqrt(1/8(sqrt3+1)^2)#
#=(sqrt3+1)/(2sqrt2)#
Using relation (1)
#sin33=sin18cos15+cos18sin15#
#=(sqrt5-1)/4*(sqrt3+1)/(2sqrt2)+sqrt(10+2sqrt5)/4*(sqrt3-1)/(2sqrt2)#
#=1/(8sqrt2)((sqrt5-1)(sqrt3+1)+(sqrt(10+2sqrt5))(sqrt3-1))#
