若tanα=√3,求cos2α-cos^2α/sin^2α-1

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/06 01:54:20
若tanα=√3,求cos2α-cos^2α/sin^2α-1
xQAJ0NORp"e ѕZ@DTT+v Eo0?uav{42Nn:WL:/ļb8RPe4;XNfq*΅DNVbC71x%A8ö#>yG>b/u"YS\C7 p0#HlLQ/YK-,@@:kfe3M{=m޻5֔1=h^~ M,oh}C.~yol

若tanα=√3,求cos2α-cos^2α/sin^2α-1
若tanα=√3,求cos2α-cos^2α/sin^2α-1

若tanα=√3,求cos2α-cos^2α/sin^2α-1
cos2a-1-cot^2a=2cos^2a-1/3
=2*1/(1+tan^2a)-1/3=2/(1+3)-1/3
=1/2-1/3=1/6
用到公式:
cos^2a=1/(1+tan^2a)

[cos2a-(cosa)^2]/[(sina)^2-1]
=[(cosa)^2-(sina)^2-(cosa)^2]/-(cosa)^2
=-(sina)^2/-(cosa)^2=(tana)^2=3

为了书写方便,我分子,分母分别写了
cos2a-(cosa)^2=cosa^2-sina)^-(cosa)^2=-(sina)^2
sina^2-1=-(cosa)^2
cos2α-cos^2α/sin^2α-1
=-tana^2=-3

cos2a-1-cot^2a=2cos^2a-1/3
=2*1/(1+tan^2a)-1/3=2/(1+3)-1/3
=1/2-1/3=1/6
用到公式:
cos^2a=1/(1+tan^2a)