1楼:匿名用户
1)(∵x∈(抄
πbai/2,3πdu/4),∴
x-πzhi/4∈(π/4,π/2)
∵cos(x-π/4)=√dao2/10
∴sin(x-π/4)=√[1-cos2(x-π/)]=7√2/10∴sinx=sin[(x-π/4)+π/4]=sin(x-π/4)cosπ/4+cos(x-π/4)sinπ/4=7√2/10*√2/2+√2/10*√2/2=4/5
(2)cosx=-3/5
∴sin2x=2sinxcosx=2*4/5*(-3/5)=-24/25
cos2x=1-2sin2x=-7/25
∴sin(2x+π/3)=sin2xcosπ/3+cos2xsinπ/3
=-24/25*1/2-7/25*√3/2=-(24+7√3)/50
已知cos(x-π/4)=√2/10,x∈(π/2,3π/4),1.求sinx 2.求sin(2x+π/3)的值
2楼:随缘
1)(∵x∈(π
/2,3π/4),∴x-π/4∈(π/4,π/2)∵cos(x-π/4)=√2/10
∴sin(x-π/4)=√[1-cos2(x-π/)]=7√2/10∴sinx=sin[(x-π/4)+π/4]=sin(x-π/4)cosπ/4+cos(x-π/4)sinπ/4=7√2/10*√2/2+√2/10*√2/2=4/5
(2)cosx=-3/5
∴sin2x=2sinxcosx=2*4/5*(-3/5)=-24/25
cos2x=1-2sin2x=-7/25
∴sin(2x+π/3)=sin2xcosπ/3+cos2xsinπ/3
=-24/25*1/2-7/25*√3/2=-(24+7√3)/50
已知cos(x-π/4)=√2/10,x∈(π/2,3π/4),求(1)sinx的值,(2)求sin(2x+π/3)的值。
3楼:随缘
∵x∈(π
制/2,3π/4),
∴x-πbai/4∈du(π/4,π/2)∵zhicos(x-π/4)=√dao2/10∴sin(x-π/4)=√[1-cos2(x-π/4)]=7√2/10
∴sinx=sin[(x-π/4)+π/4]=sin(x-π/4)cosπ/4+cos(x-π/4)sinπ/4=7√2/10*√2/2+√2/10*√2/2=7/10+1/10=4/5
(2)∵x∈(π/2,3π/4),
∴cosx=-3/5
∴sin2x=2sinxcosx=-24/25cos2x=1-2sin2x=-7/25
∴sin(2x+π/3)=sin2xcosπ/3+cos2xsinπ/3
=(-24/25)*1/2-7/25*√3/2=-(24+7√3)/50
已知cos(x-兀/4)=√2/10 x∈(兀/2,3兀/4)求sinx
4楼:手机用户
^^cos(x-兀/4)=cos x cos (兀/4)+sin x sin (兀/4)=√2/2 [cos x +sin x]=√2/10
所以 cos x +sin x =1/5 又 cos^2 x +sin^2 x = 1
所以 (1/5 - sin x)^2 + sin^2 x = 1, 所以 2sin^2 x - 2/5 sin x - 24/25 = 0
所以 sin x = - 4/5 (x属于(兀/2,3兀/4), sin x <0 cos x <0 ) cos x = - 3/5
sin2x=2sinxcosx=24/25, cos2x=2cos^2 x -1= - 7/25
sin(2x+兀/3)=sin 2x cos(兀/3) + cos 2x sin(兀/3)=1/2 sin 2x + √3/2 cos 2x
= 1/2 * 24/25 - √3/2 * 7/25 = (24 - 7√3)/50
已知cos(x-派/4)=根号2/10,x属于(派/2,3派/4)求sinx的值.
5楼:左右鱼耳
^cos(x-πcopy
bai/4)=cosxcosπ/4+sinxsinπ/4=(√2/2)(cosx+sinx)=√2/10
cosx+sinx=1/5
(cosx+sinx)^du2=(cosx)^2+2cosxsinx+(sinx)^2
=1+2sinxcosx
=1/25
sinxcosx=-12/25
∵x∈(π/zhi2,3π/4)
∴sinx>0
则daocosx<0
(sinx-cosx)^2=(cosx)^2+2cosxsinx+(sinx)^2-4cosxsinx
=(cosx+sinx)^2-4×(-12/25)=49/25
sinx-cosx>0
sinx-cosx=7/5
sinx=(1/5+7/5)÷2=4/5
6楼:小林爸爸
解:因为cos(x-∏/4)=√2/10
所以√2/2cosx+√2/2sinx=√2/10又因为(cosx)2+(sinx)2=1
所以得sinx=4/5
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