三角函数求值问题
tan70cos10(√3tan20-1) 注:1、题目中的70,10,20均为角度 2、根号下只有3。
tan70cos10(3tan20-1) =cot20cos10(tan60tan20-1) =cot20cos10(sin60sin20-cos60cos20)/(cos60cos20) =cot20cos10(-cos80)/(cos60cos20) =cos20/sin20*sin80(-cos80)/(cos20/2) =-2sin80cos80/sin20 =-sin160/sin20 =-sin20/sin20 =-1.
答:(cos75°)^2+(cos15°)^2+cos75°+cos15° =(cos75°)^2+(sin75°)^2+cos75°+sin75° =1+(√6-...详情>>
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