Question

计算:sin5π/6－cos2π/3＋tan（－π/4）

Answer 1

问：计算:sin5π/6－cos2π/3＋tan（－π/4）

答：亲，答案是(√3-3)/2。

答：因为sin5π/6 = sin(π/6+π/2) = cos(π/6) = √拿唤陵3/2。cos2π/3 = cos(π/3+π/3) = cos(π/3) = 1/2。tan(-π/4) = -1。故sin5π/6－cos2π/3＋tan（－消戚π/4）= √3/2 - 1/2 - 1 = (√3-3)/2。答案为链段(√3-3)/2。

答：相关拓展，需要根据sina=1/3和a在-3Π/2到Π之间两个条件判断cosa的正负情况哦~

问：可以快点不

答：跟我发的图片上结果是一样的哦

答：3Π/2到2Π之间，cosa也是大于0的呢