化简sinπ/12-√3cosπ/12化简sinπ/12-√3cosπ/12

来源：学生作业帮助网编辑：作业帮时间：2024/05/10 11:57:19

化简sinπ/12-√3cosπ/12化简sinπ/12-√3cosπ/12
化简sinπ/12-√3cosπ/12
化简sinπ/12-√3cosπ/12

化简sinπ/12-√3cosπ/12化简sinπ/12-√3cosπ/12
sinπ/12-√3cosπ/12
=2*（1/2*sinπ/12-√3/2*cosπ/12）
=2*（cosπ/3*sinπ/12-sinπ/3*sinπ/12）
=2*sin(π/12-π/3)
=-2sinπ/4
=-√2

sinπ/12-√3cosπ/12=2（1/2sinπ/12-√3/2cosπ/12）=
2（sinπ/12cosπ/3-sinπ/3cosπ/12）=2sin（π/12-π/3）=2sin-π/4=-2*√2/2=-√2

sinπ/12-√3cosπ/12
=2(1/2*sinπ/12-√3/2*cosπ/12)
=2(sinπ/12cosπ/6-cosπ/12sinπ/6)
=2sin(π/12-π/6)
=2sin(-π/12)
=-2sinπ/12

sinπ/12-√3cosπ/12
=2（1/2*sinπ/12-√3/2cosπ/12)
=2sin(π/12-π/3)
=-2sinπ/4
=-根号2

化简sinπ/12-√3cosπ/12化简sinπ/12-√3cosπ/12 sinπ/12-√3cosπ/12的值 √3cosπ/12-sinπ/12= 若方程12x²+πx-12x=0的两的根分别是α,β,则cosαcosβ-√3sinαcosβ-√cosαsinβ-sinαsinβ= 化简：1,cosπ/8的4次方- sinπ/8的4次方 2,sinπ/12-√3cosπ/12 (1+sinα+cosα)*[sin(α/2)-cos(α/2)]/√(2+2cosα)化简（3/2*π 化简 cos （π/12）+√3sin（π/12）= （sinπ/24)(cosπ/24)(cosπ/12) （sinπ/24)(cosπ/24)(cosπ/12) sinπ/12*cosπ/12=? 求值：sin(π/12)+cos(π/12) sin(π/12)-cos(π/12)= cos*π/12—sin*π/12 cosπ／12乘以sinπ／12 计算cosπ/12*sinπ/12 已知α=7/12π,那么cosα√（1-sinα）/(1+sinα)+sinα√（1-cosα）/(1-cosα)= 怎样化简函数 cosπ/4cosπ/3-sinπ/4sinπ/3cosπ/4cosπ/3-sinπ/4sinπ/3 化简√1-2sin(π-3)cos(π+3)