已知cos(a+b)=4/5 cos(a-b)=-4/5. 3派/2<a+b<2派,派/2<a-b<派 求cos2a. Cos2b
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.已知Cos(a+b)=4\/5,cos(a-b)= -4\/5,3\/2π<a+b<2π,π\/2<a-b<π,求C...
3π\/2<a+b<2π,且cos(a+b)=4\/5,则:sin(a+b)=-3\/5 π\/2<a-b<π,且cos(a-b)=-4\/5,则:sin(a-b)=3\/5 cos(2a)=cos[(a+b)+(a-b)]=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)=(4\/5)×(-4\/5)-(-3\/5)×(3\/5)=-7\/25 ...
已知cos(a+b)=4\/5,cos(a-b)=-4\/5,π\/2<a-b<π,3π\/2<a+b<2π。
∴sin(a-b)=3\/5 cos2b=cos[(a+b)-(a-b)]=cos(a+b)cos(a-b)+sin(a+b)sin(a-b)=4\/5(-4\/5)-3\/5(3\/5)=-1 (2) ∵π\/2<a-b<π,3π\/2<a+b<2π ∴π<a<3π\/2 -3π\/2<-a<-π 0<b<π 0<2b<2π 2b=π ∴ b=π\/2 ...
已知COS(a+b)=4\/5,cos(a-b)=-4\/5, 3\/2π小于a+b小于2π,π\/2小
所以sin(a+b)0,sin(a-b)0 值=(4\/5)*(-4\/5)-(3\/5)*(-3\/5)=-7\/25 ②.第二题b是π\/2bπ吧?tana+tanb=5\/6 tana*tanb=1\/6 tan(a+b)=(tana+tanb)\/(1-tana*tanb)=5\/6\/(1-1\/6)=1 a+b=π\/4(4\/π)③.x属于(0,π\/4)0π\/4-xπ\/4 所以cos(π\/4-x)0 ...
已知cos(a+b)=4\/5,cos(a-b)=-4\/5,3π\/2
3π\/2<a+b<2π,π 2<a-b<π 所以sin(a+b)<0 sin(a-b)>0 由(sinx)^2+(cosx)^2=1 所以sin(a+b)=-3\/5,sin(a-b)=3\/5 cos2b=cos[(a+b)-(a-b)]=cos(a+b)cos(a-b)+sin(a+b)sin(a-b)=-16\/25-9\/25 =-1<\/a+b ...
已知cos(α+β)=4\/5,cos(α-β)=-4\/5,3π\/2<α+β<2Π,Π\/2<α-β<...
sin(a+β)=3\/5 sin(a-β)=3\/5 cos2β =cos[(a+β)-(a-β)]=cos(a+β)cos(a-β)+sin(a+β)sin(a-β)=4\/5*(-4\/5)+3\/5*3\/5=-16\/25+9\/25 =-7\/25 cos2β=2cos²β-1 cos²β=(cos2β+1)\/2=(-7\/25+1)\/2=18\/25\/2=8\/25 cosβ=(-2√2...
已知cos(a+β)=4\/5,cos(a-β)=-4\/5,a+β(3π\/2,2π),a-β(π\/2,π...
因为cos(a+β) = 4\/5 cos(a-β)=-4\/5 且a+β(3π\/2,2π),a-β(π\/2,π),sin(a+β)= -3\/5, sin(a-β)=3\/5 由cos2a = cos(a+β+a-β)=cos(a+β)cos(a-β)-sin(a+β)sin(a-β)=-7\/25 如果帮助到你,请点击满意按钮来支持我哦 你的支持是我最大的...
已知cos(α+β)=4\/5,sin(α-β)=3\/5,且3π\/2<α+β<2π,π\/2<α-β<...
cos2β=cos[α+β-(α-β)]=4\/5*(-4\/5)+(-3\/5)*3\/5=-1 β=π-π\/4
“已知cos(α+β)=4\/5,cos(α-β)=-4\/5,α+β∈(7\/4π,2π),α-β...
cos(α+β)=4\/5,cos(α-β)=-4\/5,α+β∈(7\/4π,2π),α-β∈(3\/4π,π)sin(α+β)=-3\/5,sin(α-β)=3\/5,cos2α=cos[(α+β)+(α-β)]=cos(α+β)cos(α-β)-sin(α+β)sin(α-β)=4\/5*(-4\/5)-(-3\/5)*(3\/5)=-7\/25 ...
已知a,B属于(0,兀\/2),cosa=4\/5,cos(a+B)=-4\/5求sinB
cosa=4\/5 sina=3\/5 cos(a+B)=-4\/5 sin(a+B)=3\/5 sinB=sin[(a+B)-a]=sin(a+B)cosa-cos(a+B)sina =3\/5*4\/5+4\/5*3\/5 =24\/25
已知cos(A+B)=4\/5,cos(A-B)= -4\/5求cosAcosB的值?
cos(A+B)=cosAcosB-sinAsinB=4\/5 cos(A-B)= cosAcosB+sinAsinB=-4\/5 所以上面两个式子相加 得2cosAcosB=0 cosAcosB=0,9,