cos2x=-1/9,则tan平方x乘sin平方x= tan10度tan20度+根号3(tan10度+tan20度)=
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2个回答

1.cos2x=-1/9

2cos^2x-1=1/9 cos^2x=5/9

1-2sin^2x=1/9 sin^2x=4/9

tan平方x乘sin平方x=sin^2x*sin^2x/cos^2x=(4/9)*(4/9)/(5/9)=16/45

2.tan10度tan20度+根号3(tan10度+tan20度)

=(sin10°sin20°)/(cos10°cos20°)+√3(sin10°/cos10°+sin20°/cos20°)

=[sin10°sin20°+√3(sin(10°+20°)]/(cos10°cos20°)

=[-(1/2)cos30°+(1/2)cos10°+√3/2]/(cos10°cos20°)

=(1/2)(cos10°+√3/2)/(cos10°cos20°)

=(1/2)(cos10°+cos30°)/(cos10°cos20°)

=(cos10°cos20°)/(cos10°cos20°)

=0

我写详细些,不然会少好多步骤,

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