求「2sin50度+sin10度(1+(√3)·tan10度)」√(2sin^80
解:
∵cos10+√3sin10=2[(1/2)cos10+(√3/2)sin10]
=2(cos60cos10+sin60sin10)=2cos(60-10)=2cos50
∴{2sin50+[1+(√3)tan10]sin10}×√[2(sin80)^]
={2sin50+[1+(√3)(sin10/cos10)]sin10}×√[2(sin80)^]
={2sin50+[1+(√3)(sin10/cos10)]sin10}×√[2(sin80)^]
={2sin50+[cos10+(√3)(sin10)]sin10/cos10}×√[2(sin80)^]
=[2sin50+(2co...全部
解:
∵cos10+√3sin10=2[(1/2)cos10+(√3/2)sin10]
=2(cos60cos10+sin60sin10)=2cos(60-10)=2cos50
∴{2sin50+[1+(√3)tan10]sin10}×√[2(sin80)^]
={2sin50+[1+(√3)(sin10/cos10)]sin10}×√[2(sin80)^]
={2sin50+[1+(√3)(sin10/cos10)]sin10}×√[2(sin80)^]
={2sin50+[cos10+(√3)(sin10)]sin10/cos10}×√[2(sin80)^]
=[2sin50+(2cos50sin10/cos10)]×(√2)(sin80)
=2[(cos10sin50+cos50sin10)/cos10]×(√2)cos10
=2(sin60/cos10)×(√2)(sin80)
=(2√2)×(√3)/2=√6。
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