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
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
用诱导公式和三角恒等变换倒角,
(2cos10-sin20)/sin70
=(2cos10-cos70)/sin70
=(2cos10-(cos60cos10-sin60sin10))/sin70
=(1.5cos10+√3 /2 sin10)/sin70
=√3sin(10+60) /sin70
=√3