1/cos50度+tan10度=??
答案:2 悬赏:30
解决时间 2021-01-19 00:23
- 提问者网友：了了无期
- 2021-01-18 18:05
1/cos50度+tan10度=??
最佳答案
- 二级知识专家网友：毛毛
- 2021-01-18 19:18
1/cos50°+tg10°=1/sin40° + sin10°/cos10°
=1/sin40° +(sin10°*2sin10°) /(cos10°*2sin10°)
=1/sin40° +2(sin10°)^2 /sin20°
=1/sin40° +(1-cos20°) /sin20°
=1/sin40°+(1-cos20°)*2cos20° /sin20°*2cos20°
=[1+2cos20°-2(cos20°)^2] /sin40°
=(2cos20°-cos40°) /sin40°
=(cos20°+2sin30°sin10°) /sin40°
=(sin70°+sin10°) /sin40°
=[sin(40°+30°)+sin(40°-30°)] /sin40°
=2sin40°cos30°/sin40°=2cos30°=√3
=1/sin40° +(sin10°*2sin10°) /(cos10°*2sin10°)
=1/sin40° +2(sin10°)^2 /sin20°
=1/sin40° +(1-cos20°) /sin20°
=1/sin40°+(1-cos20°)*2cos20° /sin20°*2cos20°
=[1+2cos20°-2(cos20°)^2] /sin40°
=(2cos20°-cos40°) /sin40°
=(cos20°+2sin30°sin10°) /sin40°
=(sin70°+sin10°) /sin40°
=[sin(40°+30°)+sin(40°-30°)] /sin40°
=2sin40°cos30°/sin40°=2cos30°=√3
全部回答
- 1楼网友：人類模型
- 2021-01-18 20:17
1/cos50+tan10
=1/sin40+cos80/sin80
=(2cos40+cos80)/sin80
=(2sin(60-10)+sin10)/cos10
=(√3cos10-sin10+sin10)/cos10
=√3
=1/sin40+cos80/sin80
=(2cos40+cos80)/sin80
=(2sin(60-10)+sin10)/cos10
=(√3cos10-sin10+sin10)/cos10
=√3
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