数学三角函数---回答好有赏分哦已知cos(x-兀/4)=根号2/10.,x∈(兀/2,3/4兀)1) 求sinx的值2) 求sin(2x+兀/3).
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![数学三角函数---回答好有赏分哦已知cos(x-兀/4)=根号2/10.,x∈(兀/2,3/4兀)1) 求sinx的值2) 求sin(2x+兀/3).](/uploads/image/z/13709964-12-4.jpg?t=%E6%95%B0%E5%AD%A6%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0---%E5%9B%9E%E7%AD%94%E5%A5%BD%E6%9C%89%E8%B5%8F%E5%88%86%E5%93%A6%E5%B7%B2%E7%9F%A5cos%28x-%E5%85%80%2F4%29%3D%E6%A0%B9%E5%8F%B72%2F10.%2Cx%E2%88%88%EF%BC%88%E5%85%80%2F2%2C3%2F4%E5%85%80%EF%BC%891%EF%BC%89+%E6%B1%82sinx%E7%9A%84%E5%80%BC2%EF%BC%89+%E6%B1%82sin%EF%BC%882x%2B%E5%85%80%2F3%EF%BC%89.)
数学三角函数---回答好有赏分哦已知cos(x-兀/4)=根号2/10.,x∈(兀/2,3/4兀)1) 求sinx的值2) 求sin(2x+兀/3).
数学三角函数---回答好有赏分哦
已知cos(x-兀/4)=根号2/10.,x∈(兀/2,3/4兀)
1) 求sinx的值
2) 求sin(2x+兀/3).
数学三角函数---回答好有赏分哦已知cos(x-兀/4)=根号2/10.,x∈(兀/2,3/4兀)1) 求sinx的值2) 求sin(2x+兀/3).
对于第一小问:
答:因cos(x-兀/4)=根号2/10
所以sin(x-兀/4)=±7√2/10
因 x∈(兀/2,3/4兀)
所以sin(x-兀/4)=7√2/10
所以得:cos(x-兀/4)=√2/10=√2/2·conX+√2/2·sinX ①
sin(x-兀/4)=√2/10=√2/2·sinX-√2/2·cosX ②
①式+②式得:sinX=4/5
第二小问:由⑴知cosX=-3/5
sin2X=2sinXcosX=-24/25
cos2X=1-2sin2X=-7/25
sin(2x+兀/3)=sin2X·cosπ/3+cos2X·sinπ/3
=-24/25*√3/2+(-7/25*1/2)
=-(24+7√3)/50
1)因为π/2
sinx=cos(x-π/2)
=cos(x-π/4-π/4)
=cos(x-π/4)cos(π/4)+sin(x-π/4)sin(π/4)
=√2/10*√2/2+7√2/10*√2/2
=4/5
cosx...
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1)因为π/2
sinx=cos(x-π/2)
=cos(x-π/4-π/4)
=cos(x-π/4)cos(π/4)+sin(x-π/4)sin(π/4)
=√2/10*√2/2+7√2/10*√2/2
=4/5
cosx=-√[1-(4/5)^2]=-3/5
2)因为π/2
sin(2x+π/3)=sin2xcos(π/3)+cos2xsin(π/3)
=2sinxcosxcos(π/3)+[2(cosx)^2-1]*sin(π/3)
=2*(4/5)*(-3/5)*(1/2)+[2(-3/5)^2-1]*(√3/2)
=-12/25-7√3/50
=-(24+7√3)/50
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