已知cosA=12 13,A∈(3π 2,2π),则cos(A π 4)为
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![已知cosA=12 13,A∈(3π 2,2π),则cos(A π 4)为](/uploads/image/f/4219997-5-7.jpg?t=%E5%B7%B2%E7%9F%A5cosA%3D12+13%2CA%E2%88%88%283%CF%80+2%2C2%CF%80%29%2C%E5%88%99cos%28A+%CF%80+4%29%E4%B8%BA)
(sina+cosa)^2=1+2sinacosa=1+4/5=9/5又因为根号下cos^2a=-cosa,所以cosa
(sina+cosa)²=sina²+cosa²+2sina*cosa=(√5/2)²=5/4已知:sina²+cosa²=1可得:sina
(sina-cosa)²=(7/13)²展开整理得:1+2sinacosa=49/169解得:sinacosa=60/169(sina+cosa)²=(sina-cosa
(1)向量a*b=2cosA*sinA-2sinA*cosA=0,则向量a垂直向量b.(2)向量x*y=-ka^2+(t^2-3)t*b^2=-4k+(t^2-3)t=0,所以k=(t^2-3)t/4
|a+b|^2=|a|^2+|b|^2+2a·b=1+2-2√2sinα+sinα^2+cosα^2+2cosα(√2-sinα)+2sinαcosα=4+2√2(cosα-sinα)=4+4cos(
a∈(0,π/3)∴a+π/6∈(π/6,π/2)∴cos(a+π/6)>0∴cos(a+π/6)=√[1-sin²(a+π/6)]=√(1-16/25)=3/5∴cosa=cos[(a+π
tana=3->sina=正负1/根号10cosa=正负3/根号10二者同正同负,代入(3sina-2cosa)/(2sina+5cosa)=-3/17cos^2a=9/10sina*cosa=3/1
1)sinA+cosA=1/5.(1)(sinA+cosA)^2=1/251+2sinAcosa=1/25sinAcosA=-12/25.(2)(sinA-cosA)^2=1-2sinAcosA=1-
因为sina*cosa=60/169=5/13*12/13所以sina=5/13与cosa=12/13或sina=12/13与cosa=5/13
移项!sinA-cosA/sinA+cosA=1/33sinA-3cosA=sinA+cosA2sinA=4cosA∵tanA=sinA/cosA∴tanA=2
sina+cosa=2/3(0)(sina+cosa)^2=4/9(sina)^2+(cosa)^2+2sinacosa=4/91+2sinacosa=4/92sinacosa=4/9-12sinac
(a-c)*b=(cosa-√2)*(√2-sina)=√2(sina+cosa)-sinacosa-2设(sina+cosa)=t,则可得:sinacosa=(t^2-1)/2,且t∈(-1,√2]
(a+c)·b=(0,sina-1)·(1+cosa,sina)=0+sin²a-sina=sin²a-sina令t=sina,t∈[-1,1],则(a+c)·b=t²-
sina+cosa=√2/2同平方:sin^2a+cos^2a+2sinacosa=1/21+2sinacosa=1/2-2sinacosa=1/21-2sinacosa=1+1/2=3/2sin^2
a∈(π,2π)a/2∈(π/2,π)sin(a/2)=4/5cos(a/2)=-3/5sina=2sin(a/2)cos(a/2)=2*4/5*(-3/5)=-24/25cosa=1-2sin
a∈(π,2π)a/2∈(π/2,π),a/2在第二象限所以cos(a/2)
1.(sina+cosa)/(2sina-cosa)=(tana+1)/(2tana-1)=(3+1)/(6-1)=4/52.tana=sina/cosa=3sina=3cosasin²a+
因为a垂直于b,所以sina+2cosa=0.所以tana=-2所以sina=2/根号5,cosa=-1/根号5所以sina-cosa=2/根号5+1/根号5
由(1/sina+1/tana)×(1-cosa)/cosa=2得(1+cosa)/sina×(1-cosa)/cosa=(1-cos²a)/(sinacosa)=1/cosa=2所以cos
(sina+cosa)^2=1+2sinacosa=5/3因为a为锐角所以sina>0且cosa>0所以sina+cos=(根号15)/3