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化简sin40(tAn10%√3)

是这一步吗=(sin10/cos10-sin60/cos60)sin40 =sin40(sin10cos60-cos10sin60)/(cos10cos60) ?这是括号里通分,分母是cos10和cos60所以同分后分母是cos10cos60

(1) 1/sin10度-根号3/sin80度=(sin80°-根号3sin10)/(sin10°*sin80°)=(sin80°-2*sin60°sin10)/(sin10°*sin80°)=(sin80°-cos50° cos70°)/(sin10°*sin80°)=(sin80°-sin40° sin20°)/(sin10°*sin80°)=(2cos60°sin20 sin20)/(sin10°*sin80°)=2sin20°/(

解:sin40°(tan10°-√3)=sin40°(sin10°/cos10°-√3)=sin40°(sin10°-√3cos10°)/cos10°=2sin40°[sin10°cos60°-cos10°sin60°]/cos10°=2sin40*sin(-50°)/cos10°=-2sin40°cos40°/cos10°=-sin80°/cos10°=-1

sin40(tan10-√3)=sin40*(tan10-tan60)=sin40*(sin10*cos60-sin60cos10)/cos10cos60=-sin40*sin50/(1/2*cos10)=-sin40*cos40/(1/2*cos10)=-1/2*sin80/(1/2*cos10)=-cos10/cos10=-1

sin40(tan10+√3)=sin40(sin10/cos10 + √3)= (sin40/cos10)(sin10 + √3 cos10)= 2(sin40/cos10)[(1/2)sin10 + (√3 /2)cos10]= 2(sin40/cos10)(cos60sin10 + sin60cos10)= 2(sin40/cos10)sin(10 + 60)= 2sin40*sin70/cos10= 2sin40*sin70/sin80= 2sin40*sin70/[2sin40cos40]= sin70/cos40不能再化简了, 所以题目本身有疑点别忘记采纳为你认真回答问题的人不要关闭问题

原式=(tan10-tan60)sin40 =(sin10/cos10-sin60/cos60)sin40 =sin40(sin10cos60-cos10sin60)/(cos10cos60) =sin40*sin(10-60)/(sin80sin30)) =-sin40*sin50/(sin80sin30)) =-sin40cos40/(2sis40cos40sin30) =-1/(2sin30)=-1

sin40°(√3-tan10°) =sin40°(√3-sin10°/cos10°) =sin40°(√3cos10°-sin10°)/cos10° =2sin40°(√3/2cos10°-1/2sin10°)/cos10° =2sin40°sin(60°-10°)/cos10° =2sin40°sin50°/cos10° =2sin40°cos40°/sin80° =sin80°/sin80° =1

sin40'*[tan10'-3^(1/2)]=sin40'*(tan10'-tan60') =sin40'*(sin10'cos60'-sin60'cos10')/(cos10'cos60') =-sin40'sin50'/(cos10'cos60')=-2sin40'cos40'/cos10'=-sin80'/sin80'=-1 '表示度(为了方便)

原式=sin40°(tan10°-tan60°)=sin40°(sin(10°-60°)/(cos10°cos60°))=2sin40°sin(-50°)/cos10°=-2sin40°cos40°/sin80°=-sin80°/sin80°=-1

sin40°(tan10°-根号3)=sin40*sin10/cos10-根号3*sin40=sin40*cos80/cos10-根号3*sin40=sin40*cos80/2sin40*cos40-根号3*sin40=(1/2)*cos80/cos40-根号3*sin40=[(1/2)*cos80-根号3sin40*sin40]/cos40=[(1/2)cos80-根号3/2*sin80]/cos40=(sin30*cos80-cos30*sin80)/cos40=sin(30-80)/cos40=-sin50/cos40=-1

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