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※ 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 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 122.125.39.189
veryhard :謝謝指教 08/02 22:34