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**HallsofIvy** You have sin(40)= sin(4(10)), sin(80)= sin(8(10)), and sin(120)= sin(12(10)).

So I would recommend some trig identities:

sin(x+ y)= sin(x)cos(y)+ cos(x)sin(y)

cos(x+ y)= cos(x)cos(y)- sin(x)sin(y)

Setting x= y= t, sin(2t)= 2 sin(t)cos(t) and cos(2t)= cos^2(t)- sin^2(t).

sin(3t)= sin(2t+ t)= sin(t)cos(2t)+ cos(t)sin(2t)= sin(t)(cos^2(t)- sin^2(t))+ cos(t)(2sin(t)cos(t))= sin(t)cos^2(t)- sin^3(t)+ 2sin(t)cos^2(t)= 3sin(t)cos^2(t)- sin^3(t).

With t= 40, sin(80)= 2 sin(40)cos(40) and sin(120)= 3 sin(40)cos^2(40)- sin^3(40).

With t= 10, sin(20)= 2sin(10)cos(10) and cos(20)= cos^2(10)- sin^2(10).

sin(40)= 2sin(20)cos(20)= 2sin(10)cos(10)(cos^2(10)- sin^2(10))= 2sin(10)cos^3(10)- 2sin^3(10)cos(10) and

cos(40)= cos^2(20)- sin^2(20)= (cos^2(10)- sin^2(10))^2- (2sin(10)cos(10)^2= cos^4(10)- 2cos^2(10)sin^2(10)+ sin^4(10)- 4sin^2(10)cos^2(10)=

cos^4(10)- 6cos^2(10)sin^2(10)+ sin^4(10)

Keep going!