sin(π-3)<sin(π/2-2)为什么
sin(\u03c0-2)+sin(3\u03c0+2)=sin(\u03c0-2)+sin(3\u03c0+2)=
sin2-sin2=0
\u7531\u516c\u5f0f\u53ef\u5f97\uff0csin\uff082\u03c0/3-A\uff09=sin\uff082\u03c0/3\uff09cosA-sinAcos\uff082\u03c0/3\uff09=\u221a3/2cosA-\uff08-1/2\uff09sinA=\u221a3/2cosA+\uff081/2\uff09sinA\uff0c\u4e5f\u53ef\u4ee5\u76f4\u63a5\u7528\u516c\u5f0fsina=sin\uff08\u03c0-a\uff09\u6240\u4ee5\u662fsin\uff082\u03c0/3-A\uff09=sin\uff08\u03c0-2\u03c0/3+A\uff09=sin\uff08\u03c0/3+A\uff09
解:sin(π-3)=sin3因为π/2<3<π
所以sin3〉0(弧度为3的角在第二象限 )
sin(π/2-2)=cos2
因为π/2<2<π,所以cos2〈0(弧度为2的角在第二象限 )
因此sin(π-3)>sin(π/2-2)
sin(π-3)=sin0.14>0
sin(π/2-2)=-sin0.43<0
sin(π-3)>sin(π/2-2)
are you sure it is sin(π-3)<sin(π/2-sin2)?
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