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December 20, 2014

December 20, 2014

Posted by **Anonymous** on Friday, January 11, 2013 at 10:49am.

then

sin^12 θ + 3 sin^10 θ + 3 sin^8 θ+ sin^6 θ + 2 sin^4 θ + 2 sin^2 θ – 2 =?

- maths -
**Steve**, Friday, January 11, 2013 at 11:32amsince 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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