Please answer the following question
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A.Maths Trigonometry
2008-06-08 9:36 pm
回答 (2)
2008-06-08 10:06 pm
✔ 最佳答案
sin²x=1-sinx1-cos²x=1-sinx
cos²x=sinx
∴cos^4x=sin²x
∴cos^4x+cos²x-1=sin²x+sinx+1=0
From (a), cos²x-1=-cos^4x
∴cos^8x+cos^6x+cos²x-1=cos^8x+cos^6x-cos^4x
=cos^4x(cos^4x+cos²x-1)
=cos^4x(0)(From (a))
=0
2008-06-08 10:00 pm
sinθ+sinθ-1=0
sinθ=1-sinθ
1-sinθ=sinθ
cosθ=sinθ
cos4θ + cosθ - 1
=(cosθ)+cosθ-1
=sinθ+sinθ-1
=0
cos8θ + cos6θ + cosθ - 1
=(cosθ)4+(cosθ)+cosθ-1
=sin4θ+sinθ+sinθ-1
=sin4θ+sinθ-sinθ
=sinθ(sinθ+sinθ-1)
=0
sinθ=1-sinθ
1-sinθ=sinθ
cosθ=sinθ
cos4θ + cosθ - 1
=(cosθ)+cosθ-1
=sinθ+sinθ-1
=0
cos8θ + cos6θ + cosθ - 1
=(cosθ)4+(cosθ)+cosθ-1
=sin4θ+sinθ+sinθ-1
=sin4θ+sinθ-sinθ
=sinθ(sinθ+sinθ-1)
=0
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