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 =?

  1. Steve

    since sin^2 + cos^2 = 1,
    sin^2 θ = cos θ

    since cosθ + cos^2 θ = 1,
    cosθ = ±(√5-1)/2

    Just plug that in for sin^2 θ, and you have

    ((√5-1)/2)^6 + 3((√5-1)/2)^5 + 3((√5-1)/2)^4 + ((√5-1)/2)^3 + 2((√5-1)/2)^2 + 2((√5-1)/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+x-1)
    = ((√5-1)/2)^3 ((√5+1)/2)^3 + 2(x^2+x-1)

    But, since (√5-1)/2 is a root of x^2+x-1 = 0 (from our original condition),

    = (5-1)^3/64 + 2(0)
    = 1

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