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1 Cos50 tAn10

1/cos50°+tg10°=1/sin40° + sin10°/cos10° =1/sin40° +(sin10°*2sin10°) /(cos10°*2sin10°) =1/sin40° +2(sin10°)^2 /sin20° =1/sin40° +(1-cos20°) /sin20° =1/sin40°+(1-cos20°)*2cos20° /sin20°*2cos20° =[1+2cos20°-2(cos20°)^2] /sin40° =(...

解: 1/(cos50°)+tan10°-2√3 ≈1.55572382686+0.176326981-2x1.732051 =-1.73205081 ≈-√3

1

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

sec50°+tan10°=1cos50°+sin10°cos10°=2sin50°2sin50°cos50°+sin10°cos10°=2sin50°sin100°+sin10°cos10°=2sin50°cos10°+sin10°cos10°=2sin50°+sin10°cos10°=sin50°+(sin50°+sin10°)cos10°=sin50°+cos20°cos10°=cos40°+cos20°cos10°=2cos30°cos10°...

因为 tan50°=tan(60°-10°) =(tan60°-tan10°)/(1+√3tan10°) sin50*(1+tan60*tan10) =cos50°*(tan60°-tan10°)*(1+√3*tan10)/(1+√3tan10°) =cos50°(tan60°-tan10°) =cos50°(sin60°/cos60°-sin10°/cos10°) =cos50° (sin60°cos10°-sin10°cos60°)/(cos...

求值:sin50度(1+根号3倍tan10度) =sin50(1+tan60tan10) =sin50(tan60-tan10)/tan(60-10) =cos50(tan60-tan10) =cos50(sin60cos10-sin10cos60)/cos60cos10 =cos50sin50/cos60cos10 =cos10/(2*1/2*cos10) =cos10/cos10 =1

sin50º×(1+√3×tan10º) =sin50º×(1+√3×sin10º/cos10º) =sin50º×(cos10º+√3×sin10º)/cos10º =sin50º×2(1/2cos10º+√3/2×sin10º)/sin80º =sin50º×2(cos60ºc...

因为 tan50°=tan(60°-10°) =(tan60°-tan10°)/(1+√3tan10°) sin50*(1+tan60*tan10) =cos50°*(tan60°-tan10°)*(1+√3*tan10)/(1+√3tan10°) =cos50°(tan60°-tan10°) =cos50°(sin60°/cos60°-sin10°/cos10°) =cos50° (sin60°cos10°-sin10°cos60°)/(cos...

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