求下列各三角函数值sin18π,cos25π,sin13π/3,cos103π/4,tan17π/4
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求下列各三角函数值sin18π,cos25π,sin13π/3,cos103π/4,tan17π/4
sin(18π)=0
cos(25π)=-1
sin(13π/3)=√3/2
cos(103π/4)=-√2/2
tan(17π/4)=1
再问: 详细过程
再答: sin(18π)=sin(9×2π+0)=sin0=0 cos(25π)=cos(12×2π+π)=cosπ=-1 sin(13π/3)=sin[4π+(π/3)]=sin(π/3)=√3/2 cos(103√/4)=cos[25π+(3π/4)]=cos(3π/4)=cos[π-π/4]=-cos(π/4)=-√2/2 tan(17π/4)=tan(4π+π/4)=tan(π/4)=1
cos(25π)=-1
sin(13π/3)=√3/2
cos(103π/4)=-√2/2
tan(17π/4)=1
再问: 详细过程
再答: sin(18π)=sin(9×2π+0)=sin0=0 cos(25π)=cos(12×2π+π)=cosπ=-1 sin(13π/3)=sin[4π+(π/3)]=sin(π/3)=√3/2 cos(103√/4)=cos[25π+(3π/4)]=cos(3π/4)=cos[π-π/4]=-cos(π/4)=-√2/2 tan(17π/4)=tan(4π+π/4)=tan(π/4)=1
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