高一数学题求解T^T 求步骤!
1个回答
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=√3/2
过程:
(1+cos20°)/2sin20°
=2cos²10°/4sin10°cos10°
=cos10°/(2sin10°)
sin10°(cot5°-tan5°)
=sin10°(cos5°/sin5°-sin5°/cos5°)
=2sin5°cos5°(cos5°/sin5°-sin5°/cos5°)
=2cos²5°-2sin²5°
=2cos10°
∴ (1+cos20°)/2sin20° — sin10°(cot5°-tan5°)
=cos10°/(2sin10°)-2cos10°
=[cos10°-2sin10°2cos10°]/(2sin10°)
=[cos10°-2sin20°]/(2sin10°)
=[cos10°-2sin(30°-10°)]/(2sin10°)
=[cos10°-2sin30°cos10°+2cos30°sin10°]/(2sin10°)
=(2cos30°)/2
=√3/2
如果您认可我的回答,请选为满意答案,并点击好评,谢谢!
过程:
(1+cos20°)/2sin20°
=2cos²10°/4sin10°cos10°
=cos10°/(2sin10°)
sin10°(cot5°-tan5°)
=sin10°(cos5°/sin5°-sin5°/cos5°)
=2sin5°cos5°(cos5°/sin5°-sin5°/cos5°)
=2cos²5°-2sin²5°
=2cos10°
∴ (1+cos20°)/2sin20° — sin10°(cot5°-tan5°)
=cos10°/(2sin10°)-2cos10°
=[cos10°-2sin10°2cos10°]/(2sin10°)
=[cos10°-2sin20°]/(2sin10°)
=[cos10°-2sin(30°-10°)]/(2sin10°)
=[cos10°-2sin30°cos10°+2cos30°sin10°]/(2sin10°)
=(2cos30°)/2
=√3/2
如果您认可我的回答,请选为满意答案,并点击好评,谢谢!
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