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sin50 13tAn10

1+√3tan10 =1+√3sin10/cos10 =(cos10+√3sin10)/cos10 =2sin(10+30)/cos10 =2sin40/cos10 sin50(1+√3tan10) =(2sin40sin50)/cos10 =[cos(50-40)-cos(50+40)]/cos10 =cos10/cos10 =1

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

解:sin50°(1+√3 tan10°)-cos20°/cos80°√(1-cos20°) =[sin50°(cos10°+√3 sin10°)/cos10°-cos20°]/√2sin²10° =(1-cos20°)/√2sin²10° =√2

sin50°(1+根号3倍tan10°) =sin50(1+tan60tan10) =sin50(tan60-tan10)/tan(60-10) =cos50(tan60-tan10) =cos50(sin60cos10-sin10cos50)/cos60cos10 =cos50sin50/cos60cos10 =sin100/2cos60cos10 =cos10/(2*1/2*cos10) =cos10/cos10 =1

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...

显然 1+√3tan10=1+tan60*tan10 =(cos60*cos10+sin60*sin10)/(cos60*cos10) =2cos(60-10)/cos10 =2cos50/cos10 于是原式 =(2sin50+2sin10*cos50/cos10) * (√2 *sin80) =2√2 *(sin50*cos10+sin10cos50) *(sin80/cos10) =2√2 *sin(50+10) =2√2 *√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°...

tan40°=tan(30°+10°)=(tan30°+tan10°)/(1-tan30°tan10°) =(√3/3 +tan10°)/[1-(√3/3)tan10°] =(1+√3tan10°)/(√3-tan10°) tan10°-√3=-(1+√3tan10°)/tan40° sin40°(tan10°-√3) =sin40°·[-(1+√3tan10°)/tan40°] =sin40°·[-(1+√3tan10°)/(sin40°/cos4...

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没有特别简单的方法,只能按部就班, 利用三角函数基本公式变形化简 [2sin50°+sin80°(1+√3tan10°)]/√(1+cos10°) =[2sin50°+cos10°(1+√3sin10°/cos10º)]/√(2cos²5°) =[2sin50°+2(1/2cos10º+√3/2sin10°)]/(√2cos5°) =[2sin50°+2(si...

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