已知tanα=3,求2/3*sin2α+1/4*cos2α

来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/24 05:28:31
已知tanα=3,求2/3*sin2α+1/4*cos2α已知tanα=3,求2/3*sin2α+1/4*cos2α已知tanα=3,求2/3*sin2α+1/4*cos2αsin2α=2tanα/(

已知tanα=3,求2/3*sin2α+1/4*cos2α
已知tanα=3,求2/3*sin2α+1/4*cos2α

已知tanα=3,求2/3*sin2α+1/4*cos2α
sin2α=2tanα/(1+tan²α)
=2*3/(1+3²)
=6/10
=3/5
cos2α=(1-tan²α)/(1+tan²α)
=(1-3²)/(1+3²)
=-8/10
=-4/5
2/3*sin2α+1/4*cos2α
=2/3*3/5+1/4*(-4/5)
=2/5-1/5
=1/5

tanα=3
2/3*sin2α+1/4*cos2α
= (2/3*sin2α+1/4*cos2α)/1
= (2/3*sin2α+1/4*cos2α)/(sin^2α+cos^2α)
= (2/3*2sinαcosα+1/4*cos^2α-1/4*sin^2α)/(sin^2α+cos^2α)
分子分母同除以cos^2α:
= (2/3*2tanα...

全部展开

tanα=3
2/3*sin2α+1/4*cos2α
= (2/3*sin2α+1/4*cos2α)/1
= (2/3*sin2α+1/4*cos2α)/(sin^2α+cos^2α)
= (2/3*2sinαcosα+1/4*cos^2α-1/4*sin^2α)/(sin^2α+cos^2α)
分子分母同除以cos^2α:
= (2/3*2tanα+1/4-1/4*tan^2α)/(tan^2α+1)
= (2/3*2*3+1/4-1/4*3^2)/(3^2+1)
= (4+1/4-9/4)/10
= (4-2)/10
= 1/5

收起