三角函数,复数

2025-05-18 22:43:30
推荐回答(1个)
回答1:

sin(A+B) = sinAcosB + cosAsinB (1)
sin(A-B) = sinAcosB - cosAsinB (2)

(1)-(2)
2cosAsinB = sin(+B) -sin(A-B)

2sin(1/2)θ (1+cosθ + cos2θ +...+ cosnθ )
=2sin(1/2)θ + 2cosθsin(1/2)θ+ ....+ 2cosnθsin(1/2)θ
=2sin(1/2)θ + [sin(3/2)θ - sin(1/2)θ]+ ....+ [ sin(n+(1/2)θ) -sin(n-(1/2)θ) ]
=2sin(1/2)θ +sin(n+(1/2)θ) - sin(1/2)θ
=sin(1/2)θ +sin(n+(1/2)θ)
=>

(1+cosθ + cos2θ +...+ cosnθ ) = 1/2 + sin(n+(1/2)θ)/ sin(1/2)θ