Question

一道高中数学题 高手进

求（2cos10°-sin20°）除以sin70°

Answer 1

这道题实际上有一个技巧
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+1/2sin10)/sin70
=根号3sin(10+60)/sin70
=根号3

Answer 2

sin20=sin(30-10)
所以 (2cos10°—sin20°)/sin70°
=(2cos10°—sin（30-10)）/sin70°
=(2cos10-(sin30cos10-cos30sin10))/sin70
=(2cos10-1/2cos10+√3/2sin10)/sin70
=(3/2sin10+√3/2sin10)/sin70
=√3（√3/2cos10+1/2sin10)/sin70
=√3sin(10+60)/sin70
=√3

Answer 3

用诱导公式和三角恒等变换倒角，

Answer 4

（2cos10-sin20）/sin70
=(2cos10-cos70)/sin70
=(2cos10-(cos60cos10-sin60sin10))/sin70
=(1.5cos10+√3 /2 sin10)/sin70
=√3sin(10+60) /sin70
=√3