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Cot10°

ctg10-4cos10 =(cos10-4sin10cos10)/sin10 =(cos10-2sin20)/sin10 =(sin80-sin20-sin20)/sin10 =(2cos50sin30-sin20)/sin10 =(cos50-sin20)/sin10 =(sin40-sin20)/sin10 =2cos30sin10/sin10 =√3

cos80=sin10=1/√(cot²10+1)

(1) ctg10-4cos10 =(cos10-4sin10cos10)/sin10 =(cos10-2sin20)/sin10 =(sin80-sin20-sin20)/sin10 =(2cos50sin30-sin20)/sin10 =(cos50-sin20)/sin10 =(sin40-sin20)/sin10 =2cos30sin10/sin10 =√3 (2) tan20+4sin20 =sin20/cos20+4sin20 =(si...

∵cot(α-10°)= 3 3 ,∴α-10°=60°,∴α=70°.故选C.

cot10-4cos10 =cos10/sin10-4cos10 =[cos10-4sin10cos10]/sin10 =(cos10-2sin20)/sin10 =[cos10-2sin(30-10)]/sin10 =(cos10-2sin30cos10+2cos30sin10)/sin10 =(cos10-2*1/2*cos10+2√3/2*sin10)/sin10 =(cos10-cos10+√3*sin10)/sin10 =√3*sin10/...

什么意思?

y=lncotx =lncosx/sinx =lncosx-lnsinx =-sinx/cosx-cosx/sinx =-(sin^2x+cosx^2)/sinxcosx =-1/sinxcosx =-2/2sinxcosx =-2/sin2x =-2csc2x.

原式=(2sin20°+cos10°+sin20°sin10°/cos20°)/(1/sin40°+cos80°/sin80°) =[(2sin20°cos20°+cos10°cos20°+sin20°sin10°)/cos20°]/(2cos40°/sin80°+cos80°/sin80°) =[(sin40°+cos10°)/cos20°]/[(2cos40°+cos80°)/sin80°] =[(sin40°+cos10°)sin80°/c...

直接利用闭合电路的欧姆定律I=U/R,得I=22√2sin(cot+30°)A

解:((1+cos20°)/2sin20°)-sin10°(cot5°-tan5°) =((1+cos20°)/4sin10°cos10°)-sin10°(cot5°-tan5°) =(2cos10°/4sin10°)-2sin5°cos5°(cot5°-tan5°) =(cos10°/2sin10°)-2((cos5°)^2-(sin5°)^2) =(cos10°/2sin10°)-2cos10° =(cos10°-4sin10°cos10...

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