求值:[cos 40°+sin 50°(1+√3tan 10°)]/[sin 70°√(1+sin 50°)].
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/18 02:17:21
求值:[cos 40°+sin 50°(1+√3tan 10°)]/[sin 70°√(1+sin 50°)].
[cos 40°+sin 50°(1+√3tan 10°)]/[sin 70°√(1+sin 50°)]
=[cos40°+sin50°(1+√3tan10°)]/[sin70°√(1+cos40°)]
=[cos40°+sin50°×(tan60°-tan10°)/tan50°]/sin70°(√2)cos20°
=[cos40°+(tan60°-tan10°)cos50°]/sin70°(√2)cos20°
=[cos40°+√3cos50°-tan10°cos50°]/ (√2)cos20°^2
=[cos40°+√3sin40°-tan10°sin40°] /√2/2(1+cos40°)
=2[(1/2)cos40+(√3/2)sin40°]-(sin10°/cos10°)sin40° /√2/2(1+cos40°)
=2(cos60°cos40°+sin60°sin40°)-[(sin10°)∧2/cos10°sin10°]sin40°/√2/2(1+cos40°)
=2cos20°-[(1-cos20°)/sin20°]2sin20°cos20°/√2/2(1+cos40°)
=2cos20°-2cos20°+2(cos20°) ^2/√2/2(1+cos40°)
=1+cos40°/√2/2(1+cos40°)
=√2
=[cos40°+sin50°(1+√3tan10°)]/[sin70°√(1+cos40°)]
=[cos40°+sin50°×(tan60°-tan10°)/tan50°]/sin70°(√2)cos20°
=[cos40°+(tan60°-tan10°)cos50°]/sin70°(√2)cos20°
=[cos40°+√3cos50°-tan10°cos50°]/ (√2)cos20°^2
=[cos40°+√3sin40°-tan10°sin40°] /√2/2(1+cos40°)
=2[(1/2)cos40+(√3/2)sin40°]-(sin10°/cos10°)sin40° /√2/2(1+cos40°)
=2(cos60°cos40°+sin60°sin40°)-[(sin10°)∧2/cos10°sin10°]sin40°/√2/2(1+cos40°)
=2cos20°-[(1-cos20°)/sin20°]2sin20°cos20°/√2/2(1+cos40°)
=2cos20°-2cos20°+2(cos20°) ^2/√2/2(1+cos40°)
=1+cos40°/√2/2(1+cos40°)
=√2
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