if 18∘is a root of cos3θ=sin2θ, and cos3θ=4cos³θ-3cosθ, show that 18∘is a root of
4sin²θ+2sinθ-1=0
double angle formula
2007-02-19 3:24 am
回答 (1)
2007-02-19 3:31 am
✔ 最佳答案
if 18∘is a root of cos3θ=sin2θ, and cos3θ=4cos³θ-3cosθ, show that 18∘is a root of 4sin²θ+2sinθ-1=0
from
4sin²θ+2sinθ-1=0
4(1-cos²θ)+2sinθ-1=0
4cos²θ-2sinθ-3=0
4cos³θ-2sinθcosθ-3cosθ=0
cos3θ=sin2θ
Since 18∘is a root of cos3θ=sin2θ
So 18∘is a root of 4sin²θ+2sinθ-1=0
2007-02-18 19:33:31 補充：
2sinθcosθ=sin2θ
收錄日期: 2021-04-25 16:52:32
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