①化简1+cos(π/2 +α)*sin(π/2 -α)*tan(π+α)
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/05/03 20:51:11
①化简1+cos(π/2 +α)*sin(π/2 -α)*tan(π+α)
②计算(1)tan(π/5)+tan(2π/5)+tan(3π/5)+tan(4π/5)
(2)sin(-60°)+cos(225°)+tan135°
(3)cos(π/5)+cos(2π/5)+cos(3π/5)+cos(4π/5)
(4)tan10°+tan170°+sin1866°-sin(-606°)
②计算(1)tan(π/5)+tan(2π/5)+tan(3π/5)+tan(4π/5)
(2)sin(-60°)+cos(225°)+tan135°
(3)cos(π/5)+cos(2π/5)+cos(3π/5)+cos(4π/5)
(4)tan10°+tan170°+sin1866°-sin(-606°)
第一问:根据诱导公式:
1+cos(π/2 +α)*sin(π/2 -α)*tan(π+α)
=1+sinαcosα×(-sinα/cosα)
=1-sin^2α
第二问:
使用和差化积公式
tan(π/5)+tan(2π/5)+tan(3π/5)+tan(4π/5)
=sin(π)/cos(π/5)×cos(4π/5)+sin(π)/cos(2π/5)×cos(3π/5)
=0
2.使用诱导公式
sin(-60°)+cos(225°)+tan135°
=-√3/2-√2/2-1
3.使用和差化积公式
cos(π/5)+cos(2π/5)+cos(3π/5)+cos(4π/5)
=2cos(π/2)cos(3π/10)+2cos(π/2)cos(π/10)
=0
4.使用和差化积公式
tan10°+tan170°+sin1866°-sin(-606°)
=sin(π)/cos10°cos170°+0
=0
祝:楼主学业进步
1+cos(π/2 +α)*sin(π/2 -α)*tan(π+α)
=1+sinαcosα×(-sinα/cosα)
=1-sin^2α
第二问:
使用和差化积公式
tan(π/5)+tan(2π/5)+tan(3π/5)+tan(4π/5)
=sin(π)/cos(π/5)×cos(4π/5)+sin(π)/cos(2π/5)×cos(3π/5)
=0
2.使用诱导公式
sin(-60°)+cos(225°)+tan135°
=-√3/2-√2/2-1
3.使用和差化积公式
cos(π/5)+cos(2π/5)+cos(3π/5)+cos(4π/5)
=2cos(π/2)cos(3π/10)+2cos(π/2)cos(π/10)
=0
4.使用和差化积公式
tan10°+tan170°+sin1866°-sin(-606°)
=sin(π)/cos10°cos170°+0
=0
祝:楼主学业进步
化简:[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(
①化简1+cos(π/2 +α)*sin(π/2 -α)*tan(π+α)
求证:[sin^2(π+α)*cos(π/2-α)+tan(2π-α)*cos(-α)]/-sin^2(-α)+tan(
化简:(2cos^2α-1)/[2tan(π/4-α)sin^2(π/4+α)]
化简(2cosα-1)/(2tan(π/4-α)*sin(π/4+α))
化简(2cos²α-1)/2tan(π/4+α)sin²(π/4-α)
化简(2cos²α-1)/[2tan(π/4-α)sin²(π/4+α)]
化简(sinα)^2tanα+cosα+(cosα)^2*1/tanα+2sinαcosα
已知tan(π-α)=1/2,则2sinα+cosα/sinα-cosα=
求证:(1-sinα+cosα)/(1+sinα+cosα)=tan(π/4-α/2)
求证:1-2sinαcosα/cos²α-sin²α=tan(π/4-α)
试证明:1+2sinαcosα/cos平方α-sin平方α=tan(π/4-α)