作业帮cos40°+sin50°×(1+√3cot10°)最好能讲讲怎么做这类题?[cos40°+sin50°×(1+√3cot80°)]/sin70°×√1+sin50°
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/06 09:26:54
cos40°+sin50°×(1+√3cot10°)最好能讲讲怎么做这类题?[cos40°+sin50°×(1+√3cot80°)]/sin70°×√1+sin50°
cos40°+sin50°×(1+√3cot10°)
最好能讲讲怎么做这类题?
[cos40°+sin50°×(1+√3cot80°)]/sin70°×√1+sin50°
cos40°+sin50°×(1+√3cot10°)最好能讲讲怎么做这类题?[cos40°+sin50°×(1+√3cot80°)]/sin70°×√1+sin50°
先化简分子：
1+√3cot80°=1+√3tan10°
=(cos10°+√3sin10°)/cos10°
=2(1/2cos10°+√3/2sin10°)/cos10°
=2(sin30°cos10°+cos30°sin10°)/cos10°
=2sin40°/cos10°
故cos40°+sin50°×(1+√3cot80°)
=cos40°+2sin50°sin40°/cos10°
=cos40°+2sin50°cos50°/cos10°
=cos40°+sin100°/cos10°
=cos40°+1
=2cos²20°
分母sin70°×√1+sin50°
=sin70°×√(sin²25°+2sin25°cos25°+cos²25°)
=sin70°×√(sin25°+cos25°)²
=sin70°(sin25°+cos25°)
=√2(cos45°sin25°+sin45°cos25°)sin70°
=√2sin70°sin70°
=√2sin²70°
=√2cos²70°
故原式=2/√2=√2
[cos40°+sin50°×(1+√3cot10°)]/sin70°×√1+sin50°
=[cos40+2sin50(1/2+√3/2*cot10)]/sin70*√(1+2sin25cos25)
=[cos40+2sin50/sin10*(sin10cos60+cos10sin60)]/sin70*(sin25+cos25)
=[cos40+2sin50sin70/s...
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[cos40°+sin50°×(1+√3cot10°)]/sin70°×√1+sin50°
=[cos40+2sin50(1/2+√3/2*cot10)]/sin70*√(1+2sin25cos25)
=[cos40+2sin50/sin10*(sin10cos60+cos10sin60)]/sin70*(sin25+cos25)
=[cos40+2sin50sin70/sin10]/sin70*√2(sin25cos45+cos25sin45)
=[cos40+2cos40cos20/sin10]/sin70*√2sin70
=√2[cos40+2cos40cos20sin20/sin20sin10]
=√2(cos40+sin80/2sin20sin10)
=……
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