发布网友 发布时间:2024-10-08 08:07
共2个回答
热心网友 时间:2024-10-08 11:42
方法一:注意70=60+10
1/2sin10-[2sin(60+10)]=1/2sin10-2(sin60cos10+cos60sin10)
=1/2sin10-√2cos10-sin10
通分
[1-2√2sin10cos10-2(sin10)^2]/2sin10=(cos20-√2sin20)/2sin10
∵cos20-√2sin20=2sin30cos20-2cos30sin20=2(sin30cos20-cos30sin20)
=2sin(30-20)=2sin10
∴原式=2sin10/2sin10=1
方法二:
1/2csc10°-2sin70
=1/(2sin10)-2sin70
=(1-4sin10sin70)/(2sin10)(用积化和差)
=[1+2(cos80-cos60)]/(2sin10)
=2cos80/(2sin10)
=2sin10/(2sin10)
=1
方法三:
1/2csc10°-2sin70
=1/(2sin10)-2sin70
=(1-4sin10sin70)/(2sin10)
=(1-4sin10sin(60+10))/2sin10
=[1-4sin10(√3/2cos10+1/28sin10)]/(2sin10)
=[1-2√3sin10cos10-2(sin10)^2]/(2sin10)
=(cos20-√3sin20)/(2sin10)
=2sin(30-20)/(2sin10)
=1
热心网友 时间:2024-10-08 11:43
解:
1/2sin10-2sin70
=1/2sin10-2sin70
=cos10/2sin10cos10-2sin70
=(cos10-2sin20sin70)/sin20
=(cos10-2cos20sin20)/sin20
=(cos10-sin40)/sin20
=(cos10-cos50)/sin20
=2sin30sin20/sin20
=1