maths
posted by Anonymous
If cosθ + cos^2 θ = 1,
then
sin^12 θ + 3 sin^10 θ + 3 sin^8 θ+ sin^6 θ + 2 sin^4 θ + 2 sin^2 θ – 2 =?

Steve
since sin^2 + cos^2 = 1,
sin^2 θ = cos θ
since cosθ + cos^2 θ = 1,
cosθ = ±(√51)/2
Just plug that in for sin^2 θ, and you have
((√51)/2)^6 + 3((√51)/2)^5 + 3((√51)/2)^4 + ((√51)/2)^3 + 2((√51)/2)^2 + 2((√51)/2)  2
= 1
Thank you wolframalpha!
Now, how can we get that for ourselves?
If you let x = sin^2 θ, the above is
x^3 (x+1)^3 + 2(x^2+x1)
= ((√51)/2)^3 ((√5+1)/2)^3 + 2(x^2+x1)
But, since (√51)/2 is a root of x^2+x1 = 0 (from our original condition),
= (51)^3/64 + 2(0)
= 1
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