化简 (21 15:57:26)（sin4α）/ [4sin2（π/4 +α）tan（π/4 -α）]=

来源：学生作业帮助网 编辑：作业帮 时间：2024/11/19 21:17:35
化简 (21 15:57:26)（sin4α）/ [4sin2（π/4 +α）tan（π/4 -α）]=
x){3 #CCS+Ss+#3{:3Lm|S_!1o7Q$t6IEXMKg~ TҴM/rtB 10 vIp[vGCΆ'>h{(V\Z`d W_ <P:.ȃDJGӵl!V A:o=Y<;h؀9 SЅ_,41'q(@DVЂX>,@4 a6zzz؂i늗+ntOha:|)XAۈb hDUq(QgS7<=Yħ3_Qx aӈ-6Mbv$@'

化简 (21 15:57:26)（sin4α）/ [4sin2（π/4 +α）tan（π/4 -α）]=
化简 (21 15:57:26)
（sin4α）/ [4sin2（π/4 +α）tan（π/4 -α）]=

化简 (21 15:57:26)（sin4α）/ [4sin2（π/4 +α）tan（π/4 -α）]=
sin(π/4+α)=cos[π/2-(π/4+α)]=cos(π/4-α)
tan(π/4-α)=sin(π/4-α)/cos(π/4-α)
所以分母=4cos²(π/4-α)*sin(π/4-α)/cos(π/4-α)
=4sin(π/4-α)cos(π/4-α)
=2sin[2(π/4-α)]
=2sin(π/2-2α)
=2cos2α
分子=sin4α=2sin2αcos2α
所以原式=sin2α

tan（π/4 -α）=cot[π/2-(π/4 -α)]=cot(π/4 +α)
所以原式=（sin4α）/ [4sin2（π/4 +α）cot(π/4 +α)]
=（sin4α）/ [4sin2（π/4 +α） *cos(π/4 +α)/sin(π/4 +α)]
=（sin4α）/ [4sin（π/4 +α） *cos(π/4 +α)]<...

全部展开

tan（π/4 -α）=cot[π/2-(π/4 -α)]=cot(π/4 +α)
所以原式=（sin4α）/ [4sin2（π/4 +α）cot(π/4 +α)]
=（sin4α）/ [4sin2（π/4 +α） *cos(π/4 +α)/sin(π/4 +α)]
=（sin4α）/ [4sin（π/4 +α） *cos(π/4 +α)]
=（sin4α）/ [2sin(π/2 +2α)]
=（sin4α）/ (2cos2α)
=2sin2αcos2α/(2cos2α)
=sin2α
分数给我吧？O(∩_∩)O谢谢！

收起

sin(15)×sin(30)×sin(75)等于多少 化简：cos(15º-α)cos15º-sin(15º-α)sin(-15º) 化简 √(2+cos2-sin^21) 注：这个是sin的平方1 化简：cos^(@+15度)+sin^(@-15度)+sin(@+180度)cos(@-180度) 化简sin(30+α)-sin(30-α) sin(a-15)=-sin(15-a)? 化简 (21 21:8:15)sin(2a+b)/sina-2cos(a+b) 化简sin(-2x) 化简sin²x 化简 sin²α sin^2(-a)化简 化简sin^2α+sin^2β+2sinαsinβcos(α+β) 化简sin(α+β)+sin(α-β)+2sinαsin(3π/2-β)= sin 15°是多少 sin sin 4.9-1/化简：[cos(a+15度)]^2+[sin(a-15度)]^2+[sin(a+180度)]*[cos(a-180度)].做出来都是1+sin(2a) 化简sin(α+β)-2sinαsinβ/2sinαsinβ+cos(α+β)