
cos(θ-π/12)=-4/5,且π/2<θ<π,求cos(2θ+π/12)的值
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cos(θ-π/12)=-4/5,sin(θ-π/12)=3/5,sin(2θ-π/6)=-24/25,cos(2θ-π/6)=7/25,
cos(2θ+π/12)=cos(2θ-π/6+π/4)=cos(2θ-π/6)cos(π/4)-sin(2θ-π/6)sin(π/4)=31根号2/50
cos(2θ+π/12)=cos(2θ-π/6+π/4)=cos(2θ-π/6)cos(π/4)-sin(2θ-π/6)sin(π/4)=31根号2/50
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