已知cos(5π/12+a)=1/3,且-《a《-π/2,则cos(π/12—a)等于?是:且-π《a《-π/2

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已知cos(5π/12+a)=1/3,且-《a《-π/2,则cos(π/12—a)等于?是:且-π《a《-π/2

已知cos(5π/12+a)=1/3,且-《a《-π/2,则cos(π/12—a)等于?是:且-π《a《-π/2
已知cos(5π/12+a)=1/3,且-《a《-π/2,则cos(π/12—a)等于?
是:且-π《a《-π/2

已知cos(5π/12+a)=1/3,且-《a《-π/2,则cos(π/12—a)等于?是:且-π《a《-π/2
因为-π0,所以5π/12+a是第四象限角,因而sin(5π/12+a)=-2√2/3.故cos(π/12—a)=cos[π/2-(5π/12+a)]=cos(π/2)*cos(5π/12+a)+sin(π/2)sin(5π/12+a)=-2√2/3.(√表示根号)