数学归纳法证明：sin²θ+sin²2θ+sin²3θ+...+sin²(nθ)=数学归纳法证明：

数学归纳法证明：sin²θ+sin²2θ+sin²3θ+...+sin²(nθ)=数学归纳法证明：
数学归纳法证明：sin²θ+sin²2θ+sin²3θ+...+sin²(nθ)=

数学归纳法证明：

数学归纳法证明：sin²θ+sin²2θ+sin²3θ+...+sin²(nθ)=数学归纳法证明：
sin^2θ=(1-cos2θ)/2
(cos2θ+cos4θ+cos6θ+.+cos2nθ)*2sin(θ)/[2sin(θ)]
=[sin(3θ)-sin(θ)+sin(5θ)-sin(3θ)+…+sin((2n+1)θ)-sin((2n-1)θ)]/[2sinθ]
=[-sin(θ)+sin((2n+1)θ)]/[2sin(θ)]
=-1/2+sin((2n+1)θ)/[2sin(θ)]=(-sin(θ)+sin((2n+1)θ)/2sin(θ)
sin(2n+1)θ-sinθ=2sin(n+1)θcosnθ
原式=(n-(sin((2n+1)θ)/[2sin(θ)]-1/2))/2=(n-2sin(n+1)θcosnθ/2sinθ)/2=(n-sin(n+1)θcosnθ/sinθ)/2
=右式 得证
sin^2θ=1-cos^2θ=1-sin2θcosθ/2sinθ
sin^2θ+sin^2(2θ)=(2-cos2θ-cos4θ)/2=(2-(cos2θ+cos4θ))/2=(2-2cos3θcosθ))/2
=(2-2cos3θcosθsinθ/sinθ))/2=(2-cos3θsin2θ/sinθ))/2
.
sin²θ+sin²2θ+sin²3θ+...+sin²(n-1)θ=(n-1-sinnθcos(n-1)θ/sinθ)/2
sin²θ+sin²2θ+sin²3θ+...+sin²(n-1)θ=(n-sin(n+1)θcosnθ/sinθ)/2