化简[sin﹙2π＋α﹚]／[cos﹙﹣π－α﹚c

化简[sin﹙2π＋α﹚]／[cos﹙﹣π－α﹚cot﹙π／2－α﹚]＝＿＿

2个回答

1.化简f(a)={[sin(π-α)cos(2π-α)cot(3π+α)]/[cot(-π-α)sin(-π-α)]}
高一三角函数化简.sin²﹙α+π﹚·cos﹙π+α﹚·cot﹙-α－2π﹚/tan﹙π+α﹚·cos²﹙-α－π﹚要
化简 [Sin(π+α)\tan(π+α)] ×[ cot(2π-α)\cos(π+α)]×[sec(2π-α)\csc
化简[sin(π/2-α）cos(π/2-α)]/cos(π+α)-[sin(π-α)cos(π/2-α)]/sin(π
化简[sin(π/2+α)*cos(π/2-α)]/cos(π+α)+[sin(π-α)*cos(π/2+α)]/sin
十万火急!化简[sin(α-π)cot(α-2π)] / [cos(α-π)tan(α-2π)]化简步骤麻烦写的详细点
化简一个三角函数式sin(3π+α）cot(-2π+α）sin(-4π+α）/tan(7π+α）cos(5π+α）能够化
sin｛π－α｝cos｛2π－α｝tan｛－α＋π｝/sin｛π＋α｝化简
化简：（Ⅰ）sin(α−2π)cos(α+π)tan(α−99π)cos(π−α)sin(3π−α)sin(−α−π)；
化简（1）tan(π−α)cos(2π−α)sin(−α+3π2)cos(−α−π)sin(−π−α)．