已知锐角α满足cosα-sinα=√5/5,则(sin2α-cos2α+1)/(1-tanα)=
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/23 03:59:28
已知锐角α满足cosα-sinα=√5/5,则(sin2α-cos2α+1)/(1-tanα)=
已知锐角α满足cosα-sinα=√5/5,则(sin2α-cos2α+1)/(1-tanα)=
已知锐角α满足cosα-sinα=√5/5,则(sin2α-cos2α+1)/(1-tanα)=
cosα-sinα=√5/5
1-2sinacosa=1/5
2sinacosa=4/5
1+2sinacosa=9/5
cosa+sina=3√5/5 (a为锐角,负值已舍)
cosa=2√5/5
sina=√5/5
tana=1/2
sin2a=2sinacosa=2 ×2√5/5× √5/5=4/5
cos2a= cos²a-sin²a=4/5-1/5=3/5
分别代入
(4/5-3/5+1)/(1-1/2)=6/5/1/2=12/5
cosα-sinα=√5/5 ,cosα=√5/5+sinα , cos^2α=1/5+2√5/5sinα+sin^2α=1-sin^2α
2sin^2α+2√5/5sinα-4/5=0 , 2(sinα+√5/10)^2-9/10=0 ,sinα=-√5/10±3√5/10 ,sinα=√5/5
cosα=2√5/5
(sin2α-cos2α+1)/(1-tanα)= ...
全部展开
cosα-sinα=√5/5 ,cosα=√5/5+sinα , cos^2α=1/5+2√5/5sinα+sin^2α=1-sin^2α
2sin^2α+2√5/5sinα-4/5=0 , 2(sinα+√5/10)^2-9/10=0 ,sinα=-√5/10±3√5/10 ,sinα=√5/5
cosα=2√5/5
(sin2α-cos2α+1)/(1-tanα)= [(cosα+sinα)^2-cos^2α+sin^2α]/(1-sinα/cosα)
=(9/5-4/5+1/5)/(1-1/2)
=12/5
收起
cosα-sinα=√5/5
1-2cosαsinα=1/5
sin2α=4/5
cos2α=3/5
cos2α=1-2sin²α
=2cos²α-1
tan²α=sin²α/cos²α
=(1-cos2α)/(1+cos2α)
=(2...
全部展开
cosα-sinα=√5/5
1-2cosαsinα=1/5
sin2α=4/5
cos2α=3/5
cos2α=1-2sin²α
=2cos²α-1
tan²α=sin²α/cos²α
=(1-cos2α)/(1+cos2α)
=(2/5)/(8/5)
=1/4
tanα=1/2
(sin2α-cos2α+1)/(1-tanα)= (4/5-3/5+1)/(1-1/2)
=(6/5)/(1/2)
=12/5
收起