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2Cos10 sin70

即在花简计算用到 sin20=sin(30-10) 所以 (2cos10°—sin20°)/sin70° =(2cos10°—sin(30-10))/sin70° =(2cos10-(sin30cos10-cos30sin10))/sin70 =(2cos10-1/2cos10+根号3/2sin10)/sin70 =(3/2sin10+根号3/2sin10)/sin70 =根号3(根号3/2cos10+1/2...

sin20=sin(30-10) 所以 (2cos10°—sin20°)/sin70° =(2cos10°—sin(30-10))/sin70° =(2cos10-(sin30cos10-cos30sin10))/sin70 =(2cos10-1/2cos10+根号3/2sin10)/sin70 =(3/2sin10+根号3/2sin10)/sin70 =根号3(根号3/2cos10+1/2sin10)/sin70 =根...

原式= (2cos10-sin20)/sin70 =(2cos10-sin20)/cos20 =[cos10+(cos10-cos70)]/cos20 =[cos10+2sin40*sin30]/cos20 =[cos10+2*1/2*sin40]/cos20 =[cos10+cos50]/cos20 =2cos30*cos20/cos20 =2cos30 = 根号3

sin20=sin(30-10) 所以 (2cos10°—sin20°)/sin70° =(2cos10°—sin(30-10))/sin70° =(2cos10-(sin30cos10-cos30sin10))/sin70 =(2cos10-1/2cos10+根号3/2sin10)/sin70 =(3/2sin10+根号3/2sin10)/sin70 =根号3(根号3/2cos10+1/2sin10)/sin70 =根...

等于√3。

解答: (2cos10°+sin20°)/sin70° =[2cos(30°-20°)+sin20°]/sin(90°-20°) =(2cos30°cos20°+2sin30°sin20°+sin20°)/cos20° =(√3cos20°+sin20°+sin20°)/cos20° =√3+2tan20° 估计是你的输入有误,应该是(2cos10°-sin20°)/sin70° 此时: (2cos10...

√(1-2sin70°cos70°)==√(sin²70°+cos²70°-2sin70°cos70°) =√(sin70°-cos70°)² =sin70°-cos70° 在直角三角形中,∠α(非直角)的对边与斜边的比叫做∠α的正弦,记作sinα,角α的邻边比斜边 叫做∠α的余弦,记作cosα。

:(2cos10º-sin20º)/sin70° =[2cos10º-sin(30º-10º)]/sin70º =[2cos10º-(sin30ºcos10º-cos30ºsin10º)]/sin70º =(2cos10º-(1/2)cos10º+(√3/2)sin10º)/sin70º =...

原式 =2sin20+[2(cos5)^2-1]+sin10sin20/cos20+1 =2sin20+cos10+sin10sin20/cos20+1 =2sin20+2cos10+(sin10sin20-cos10cos20)/cos20+1 =2sin20+2cos10-cos(20+10)/cos20+1 =2sin20cos20/cos20-cos30/cos20+2cos10+1 =sin40/cos20-sin60/cos20+2c...

sin10*2cos10*cos20*cos40*cos60/2cos10 =sin20*cos20*cos40*cos60/2cos10 =sin40*cos40*cos60/4cos10 =sin80*cos60/8cos10 =cos10/16cos10=1/16 不知道你要的是不是这个

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