Question

求证:（1+cosθ+cosθ/2） /（sinθ+sinθ/2）=sinθ/1-cosθ

求证:（1+cosθ+cosθ/2） /（sinθ+sinθ/2）=sinθ/1-cosθ

Answer 1

左边=(2cos^2θ/2+cosθ/2)/2sinθ/2cosθ/2+sinθ/2
=cosθ/2(2cosθ/2+1)/sinθ/2(2cosθ/2+1)
=cosθ/2/sinθ/2
=1/tanθ/2
=1/(1-cosθ/sinθ)
=sinθ/1-cosθ=右边
利用半角、倍角公式

Answer 2

证明：因为 [ sin θ +sin (θ/2) ] sin θ =(sin θ)^2 +sin (θ/2) sin θ,
[ 1 +cos θ +cos (θ/2) ] (1 -cos θ)
=(1 +cos θ) (1 -cos θ) +cos (θ/2) (1 -cos θ)
=1 -(cos θ)^2 +cos (θ/2) *2 [ sin (θ/2) ]^2
=(sin θ)^2 +2 sin (θ/2) cos (θ/2) *sin (θ/2)
=(sin θ)^2 +sin θ sin (θ/2),
所以 [ sin θ +sin (θ/2) ] sin θ = [ 1 +cos θ +cos (θ/2) ] (1 -cos θ).
所以 [ 1 +cos θ +cos (θ/2) ] / [ sin θ +sin (θ/2) ] = sin θ /(1 -cos θ).

= = = = = = = = =
要证 a/b =c/d,
只需证 ad =bc.