已知sin(7/6π+α)=1/3,则sin(2α-7/6π)=7/9求详解过程
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/15 15:08:54
已知sin(7/6π+α)=1/3,则sin(2α-7/6π)=7/9求详解过程
由已知得:
sin(7/6π+α)
=sin(π + 1/6π+α)
=-sin(1/6π+α)
=-sin[π/2 - (π/3 -α)]
=-cos(π/3 -α)
=-cos(α -π/3)
=1/3
那么:cos(α -π/3)=-1/3
所以:
sin(2α-7/6π)
=sin(2α-2/3π - π/2)
=-sin[π/2 - (2α-2/3π)]
=-cos(2α-2/3π)
=-[2cos平方(α-π/3) -1]
=-(2*1/9 -1)
=7/9
sin(7/6π+α)
=sin(π + 1/6π+α)
=-sin(1/6π+α)
=-sin[π/2 - (π/3 -α)]
=-cos(π/3 -α)
=-cos(α -π/3)
=1/3
那么:cos(α -π/3)=-1/3
所以:
sin(2α-7/6π)
=sin(2α-2/3π - π/2)
=-sin[π/2 - (2α-2/3π)]
=-cos(2α-2/3π)
=-[2cos平方(α-π/3) -1]
=-(2*1/9 -1)
=7/9
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