证明 (2sinα-sin2α)/(2sinα+sin2α)=tan²(θ/2)tan(π/4+α)-cot(π/4+α)=2tan2α
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/12 06:52:09
![证明 (2sinα-sin2α)/(2sinα+sin2α)=tan²(θ/2)tan(π/4+α)-cot(π/4+α)=2tan2α](/uploads/image/z/1801762-34-2.jpg?t=%E8%AF%81%E6%98%8E+%282sin%CE%B1-sin2%CE%B1%29%2F%282sin%CE%B1%2Bsin2%CE%B1%29%3Dtan%26%23178%3B%28%CE%B8%2F2%29tan%28%CF%80%2F4%2B%CE%B1%EF%BC%89-cot%EF%BC%88%CF%80%2F4%2B%CE%B1%29%3D2tan2%CE%B1)
证明 (2sinα-sin2α)/(2sinα+sin2α)=tan²(θ/2)tan(π/4+α)-cot(π/4+α)=2tan2α
证明 (2sinα-sin2α)/(2sinα+sin2α)=tan²(θ/2)
tan(π/4+α)-cot(π/4+α)=2tan2α
证明 (2sinα-sin2α)/(2sinα+sin2α)=tan²(θ/2)tan(π/4+α)-cot(π/4+α)=2tan2α
证明:左边=(2sinα-2sinαcosα)/(2sinα+2sinαcosα)=(1-cosα)/(1+cosα)
=2sin²(α/2)/2cos²(α/2)=tan²(α/2)=右边(那个是α吧)
(2sinα-sin2α)/(2sinα+sin2α)=(2sinα-2sinαcosα)/(2sinα+2sinαcosα)=(1-cosα)/(1+cosα)={1-[1-2sin²(α/2)]}/{1+[2cos²(α/2)-1]}=2sin²(α/2)/2cos²(α/2)=tan²(α/2)
从左网右推左边=(2sinα-2sinαcosα)/(2sinα+2sinαcosα)=(1-cosα)/(1+cosα)
=2sin²(α/2)/2cos²(α/2)=tan²(α/2)
证明:
(2sinα-sin2α)/(2sinα+sin2α)
= (2sinα-2sinαcosα)/(2sinα+2sinαcosα) 分子分母同除以2sinα
=(1-cosα)/(1+cosα)
=2sin²(α/2)/2cos²(α/2) ...
全部展开
证明:
(2sinα-sin2α)/(2sinα+sin2α)
= (2sinα-2sinαcosα)/(2sinα+2sinαcosα) 分子分母同除以2sinα
=(1-cosα)/(1+cosα)
=2sin²(α/2)/2cos²(α/2) 倍角公式
=tan²α/2
tan(π/4+α)-cot(π/4+α)
=tan(π/4+α)-tan[π/2-(π/4+α)]
=tan(π/4+α)-tan(π/4-α)
2tan2α
=2 tan[(π/4+α)-(π/4-α)]
=2[tan(π/4+α)-tan(π/4-α)]/[1+tan(π/4+α)tan(π/4-α)]
其中tan(π/4-α)=cot[π/2-(π/4-α)]=cot(π/4+α)=1/tan(π/4+α)
所以
2tan2α
=2[tan(π/4+α)-tan(π/4-α)]/2
=tan(π/4+α)-tan(π/4-α)
=tan(π/4+α)-cot(π/4+α)
得证
数学辅导团为您解答
收起
1、证明:
(2sinα-sin2α)/(2sinα+sin2α)=(2sinα-2sinαcosα)/(2sinα+2sinαcosα)=[2sinα(1-cosα)]/[2sinα(1+cosα)] =(1-cosα)/(1+cosα)=【2[sin(α/2)]^2】/【2[cos(α/2)]^2】=tan²(α/2)
提示; cosα=2[cos(...
全部展开
1、证明:
(2sinα-sin2α)/(2sinα+sin2α)=(2sinα-2sinαcosα)/(2sinα+2sinαcosα)=[2sinα(1-cosα)]/[2sinα(1+cosα)] =(1-cosα)/(1+cosα)=【2[sin(α/2)]^2】/【2[cos(α/2)]^2】=tan²(α/2)
提示; cosα=2[cos(α/2)]^2-1 cosα=1-2[sin(α/2)]^2
1-cosα=2[sin(α/2)]^2 1+cosα=2[cos(α/2)]^2
证明:
tan(π/4+α)-cot(π/4+α)=2tan2α
收起