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sin50 13tAn10

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...

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

显然1+根号3tan10 =1+tan60tan10 =(cos60cos10+sin60sin10)/cos60cos10 =cos(60-10)/cos60cos10 =cos50/cos60cos10 =2cos50/cos10 而1+cos20=2(cos10)^2 故得到根号(1+cos20)=根号2 *cos10 原式=(2sin50+ 2sin10*cos50/cos10) * (根号2 *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...

要用到和差化积[2sin50+sin80(1+√3tan10)]/√(1+cos10)=(2sin50+cos10(1+根号3*tan10)]/根号[2*(cos5)^2]=(2sin50+cos10+根号3*sin10)/[根号2 * cos5]=(2sin50+2sin(30+10))/[根号2 * cos5]=根号2*(sin50+sin40)/cos5=2根号2 *sin45*cos5/cos5=2

tan10-√3 =sin10/cos10-√3 =(sin10-√3cos10)/cos10 =2sin(10-60)/cos10 =-2sin50/cos10 sin40(tan10-√3) =-(2sin40sin50)/cos10 =-[cos(50-40)-cos(50+40)]/cos10 =-cos10/cos10 =-1

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...

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/...

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

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