tan70cos10(√3tan20-1)

夔钟欢3118利用三角公式,化简tan70cos10(根号3tan20 - 1) -
郟飘浩13015444486 ______[答案] 原式= 2cot20°cos10°[(√3/2)sin20°-cos20°(1/2)]/cos20° =2cos20°cos10°sin(20°-30°)/(cos20°sin20°) =2cos10°sin(-10°)/sin20° =-sin20°/sin20° =-1.

夔钟欢3118tan70cos10(√3 tan20 - 1) -
郟飘浩13015444486 ______ tan(60+20)=(tan60+tan20)/(1-tan60*tan20)=(√3+tan20)/(1-√3tan20) √3tan20-1=-(√3+tan20)/tan80 原式=tan70*cos10*[-(√3+tan20)/tan80] =[cos20/sin20]*cos10*[-(√3+tan20)*sin10/cos10] =[cos20/(2sin10*cos10]*cos10*[-(√3+tan20)*sin10/...

夔钟欢3118tan70cos10(√3tan20 - 1)化简
郟飘浩13015444486 ______ tan70cos10(√3tan20-1) =tan70*cos10(√3sin20-cos20)/cos20 =2tan70*cos10(√3sin20/2-cos20/2)/cos20 =2tan70*cos10(sin20*cos30-cos20*sin30)/cos20 =2tan70*cos10*sin(20-30)/cos20 =-2tan70*cos10*sin10/cos20 =-tan70*(2sin10*cos10)...

夔钟欢3118数学高一下红对勾问题求值:tan70+cos10(√3tan20
郟飘浩13015444486 ______ 检查以下题目:tan70后应是 “乘号”,否则只能化简成 2tan50 tan70cos10(√3*tan20-1) =tan70cos10(√3*sin20-cos20)/cos20 =tan70cos10(2sin20cos30-2cos20sin30)/cos20 =tan70cos10[-2sin(20-30)]/cos20 =-tan70[2cos10sin10]/cos20 =-tan70sin20/cos20 =-tan70tan20 =-1

夔钟欢3118[紧急求助,高中数学!!!] 化简:tan70度cos10度((根3)tan20度 - 1) 要详细步骤 辛苦 -
郟飘浩13015444486 ______ tan70度cos10度((根3)tan20度-1) =tan70°cos10°*(根3sin20°-cos20°)/cos20°=tan70°cos10°2sin(-10°)/cos20°=tan70°*(-sin20°)/cos20°=tan70°*(-tan20°)=-1

夔钟欢3118求值:tan70cos10+根号3sin10tan70 - 2cos40 -
郟飘浩13015444486 ______[答案] 原式=tan70[cos10+(根号3)sin10]-2cos40 =tan70[2(cos60cos10+sin60sin10)]-2cos40 =tan70*2cos50-2cos40 =(sin70*2cos50-2cos40*cos70)/cos70 =[2sin(70-50)]/cos70 =2sin20/cos70 =2sin20/sin20 =2

夔钟欢3118tan70°cos10°(√3tan20° - 1) -
郟飘浩13015444486 ______ tan70°cos10°(√3tan20°-1)=sin70°/cos70°*cos10°(√3sin20°/cos20°-1)=sin70°*cos10°(√3sin20°-cos20°)/cos70°cos20°=2cos20°*cos10°(√3/2sin20°-1/2cos20°)/cos70°cos20°=2*cos10°(sin60sin20°-cos20°cos20°)/cos70°=-2*cos10°*cos80°/cos70°=-2*cos10°*sin10°/cos70°=-sin20°/cos70°=-sin20°/sin20°=-1

夔钟欢3118tan70度*cos10度*(根号3*tan20度 - 1) -
郟飘浩13015444486 ______[答案] tan70度cos10度(√3 乘以tan20度-1) =tan70*cos10(√3*sin20/cos20-1) =2tan70*cos10[(√3/2)sin20/cos20-1/2] =2tan70*cos10[(√3/2)sin20-(1/2)cos20]/cos20 =2tan70*cos10[sin20cos30-cos20sin30]/cos20 =2tan70*cos10*sin(20-30)/cos20 =-2...

夔钟欢3118化简求值 tan70度cos10度根号3tan20度 -
郟飘浩13015444486 ______ 题不对吧,应该是tan70cos10(根3tan20-1)=tan70cos10【(根3sin20-cos20)/cos20】=tan70cos10·(-2sin10)/cos20(用辅助角公式)=(sin70/cos70)·(-sin20/cos20)由于sina=cos(90-a)所以原式=cos20/sin20·(-sin20/cos20)=-1

夔钟欢3118tan70*cos10*(√3tan20 - 1) 求详细过程! -
郟飘浩13015444486 ______[答案] tan70*cos10*(√3tan20-1) =tan70*cos10*(tan60*tan20-1) =tan70*cos10*[(sin60*sin20/cos60*cos20)-1] =tan70*cos10*(sin60*sin20-cos60*cos20)/(cos60*cos20) =tan70*cos10*[-cos(60+20)]/(cos60 *cos20) =-tan70*cos10*cos80/(cos60*cos20) ...