解答
化简 eiθ−1ei(k+1)θ−1
解答
(−1+cos(θ))2+sin2(θ)1−cos(θ)−cos(θ+kθ)+cos(θ+kθ)cos(θ)+sin(θ+kθ)sin(θ)+i(−1+cos(θ))2+sin2(θ)sin(θ)−sin(θ+kθ)+sin(θ+kθ)cos(θ)−cos(θ+kθ)sin(θ)
求解步骤
eiθ−1ei(k+1)θ−1
ei(k+1)θ−1=cos(kθ+θ)+isin(kθ+θ)−1
=eiθ−1cos(kθ+θ)+isin(kθ+θ)−1
eiθ−1=cos(θ)+isin(θ)−1
=cos(θ)+isin(θ)−1cos(kθ+θ)+isin(kθ+θ)−1
cos(θ)+isin(θ)−1cos(kθ+θ)+isin(kθ+θ)−1有理化:sin2(θ)+(cos(θ)−1)2sin(kθ+θ)sin(θ)+(cos(θ)−1)(cos(kθ+θ)−1)+i(sin(kθ+θ)(cos(θ)−1)−sin(θ)(cos(kθ+θ)−1))
=(cos(θ)−1)2+sin2(θ)(cos(kθ+θ)−1)(cos(θ)−1)+sin(kθ+θ)sin(θ)+i(sin(kθ+θ)(cos(θ)−1)−sin(θ)(cos(kθ+θ)−1))
将 (cos(θ)−1)2+sin2(θ)(cos(kθ+θ)−1)(cos(θ)−1)+sin(kθ+θ)sin(θ)+i(−sin(θ)(cos(kθ+θ)−1)+sin(kθ+θ)(cos(θ)−1)) 改写成标准复数形式:sin2(θ)+(−1+cos(θ))2cos(kθ+θ)cos(θ)−cos(kθ+θ)−cos(θ)+1+sin(kθ+θ)sin(θ)+sin2(θ)+(−1+cos(θ))2−cos(kθ+θ)sin(θ)+sin(θ)+sin(kθ+θ)cos(θ)−sin(kθ+θ)i
=sin2(θ)+(−1+cos(θ))2cos(kθ+θ)cos(θ)−cos(kθ+θ)−cos(θ)+1+sin(kθ+θ)sin(θ)+sin2(θ)+(−1+cos(θ))2−cos(kθ+θ)sin(θ)+sin(θ)+sin(kθ+θ)cos(θ)−sin(kθ+θ)i