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1 Cos50 tAn10

=1/cos50°+sin10°/cos10° =1/sin40°+cos80°/sin80° =2cos40°/(2sin40°cos40°)+cos80°/sin80° =(2cos40°+cos80°)/sin80° =【2cos(10°+30°)+cos80°】/sin80° =(2cos10°cos30°-2sin10sin30°+cos80°)/sin80° =(2cos30°cos10°-sin10°+cos80°)/sin...

解:

解: 这个需要科学计算器 1/cos10°+tan50°-2√3 ≈1.01543+1.19175-3.46410 =-1.25692

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°...

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

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) ...

原式=(2sin50º+cos10º+√3sin10º)/√2cos5º =[2sin50º+2(1/2*cos10º+√3/2sin10º)]/√2cos5º =(2sin50º+2cos50º)/√2cos5º=2√2(√2/2sin50º+√2/2cos50º)/√2cos5º=2√2sin95&o...

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...

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

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