→ veryhard :謝謝指教 08/02 22:34
※ 引述《veryhard (你是)》之銘言:
: Q2. 已知Θ=π/7 , 求 cos3Θ-cos2Θ+cosΘ之值. ans:1/2
z = exp(jπ/7) = cos(Θ) + j sin(Θ)
z^7 + 1 = 0
(z+1)(z^6 - z^5 + z^4 - z^3 + z^2 - z + 1) = 0
1 = (z-z^6) - (z^2-z^5) + (z^3-z^4) = Re((z-z^6) - (z^2-z^5) + (z^3-z^4))
Re(z^6) = cos(6π/7) = -cos(π/7)
Re(z) = cos(π/7), Re(z-z^6) = 2 cos(Θ)
Re(z^5) = cos(5π/7) = -cos(2π/7)
Re(z^2) = cos(2π/7), Re(z^2-z^5) = 2 cos(2Θ)
Re(z^4) = cos(4π/7) = -cos(3π/7)
Re(z^3) = cos(3π/7), Re(z^3-z^4) = 2 cos(3Θ)
cos(3Θ) - cos(2Θ) + cos(Θ) = 1/2
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