已知sin(α+β)=5sinβ,则tan(α+β)*cotα=
sin(2α+β)=5sinβ,打错了
人气:296 ℃ 时间:2020-04-28 09:25:39
解答
sin(α+β)=5sinβsinacosb+cosasinb=5sinbsinacosb=sinb(5-cosa)cosb=sinb(5-cosa)/sinacos^2b+sin^2b=1=sin^2b[(5-cosa)^2/sin^2a+1]=sin^2b[(26-10cosa)/sin^2a]sinb=+-sina/√(26-10cosa),cosb=+-(5-cosa)/√(26...sin(2α+β)=5sinβ,打错了sin(2α+β)=5sinβsin(2a)cosb+cos(2a)sinb=5sinbsin(2a)cosb=sinb[5-cos(2a)]cosb=sinb[5-cos(2a)]/sin(2a)cos^2b+sin^2b=1=sin^2b{[5-cos(2a)]^2/sin^2(2a)+1}=sin^2b[26-10cos(2a)]/sin^2(2a)sinb=sin(2a)/√[26-10cos(2a)],cosb=[5-cos(2a)]/√[26-10cos(2a)]tan(α+β)*cotα=sin(a+b)cosa/[cos(a+b)sina]=(sinacosb+cosasinb)cosa/[cosacosb-sinasinb)sina]={[sina[5-cos(2a)]/√[26-10cos(2a)]+cosasin(2a)/√[26-10cos(2a)]}cosa/{cosa[5-cos(2a)]/√[26-10cos(2a)]-sinasin(2a)/√[26-10cos(2a)]}sina={[sina[5-cos(2a)]+cosasin(2a)}cosa/{cosa[5-cos(2a)]-sinasin(2a)}sina=[5-cos(2a)+2cos^2a]cosa / {cosa[5-cos(2a)]-sinasin(2a)}=[5-cos(2a)+2cos^2a] /[5-cos(2a)-2sin^2a]=[5-(2cos^a-1)+2cos^a]/[5-(1-2sin^2a)-2sin^2a]=(5+1)/(5-1)=3/2
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