a属于(0,π 2),sina 2cosa=11 5,求tana
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解(1):∵tana=√3,π<a<3π/2∴a=π+π/3=4π/3∴cosa-sina=cos(4π/3)-sin(4π/3)=-cos(π/3)+sin(π/3)=-1/2+√3/2解(2)∵t
k=2010,所以F(x)=x^2-2ax*lnx=2ax即x-2alnx-2a=0,令左边=g(x)g(x)导函数=1-(2a)/x,令导函数为0,x=2a,列表知2a处取最小值,所以g(2a)=2
(a-x)/(x^2-x-2)>0(a-x)(x^2-x-2)>0(a-x)(x-2)(x+1)>0若a
1/2-1把2带进后面那式子算出来-1把-1再带进后面那式子算出1/2把1/2带进去又是2
cos(3π+a)=cos(2π+π+a)=COS(π+a)=-COSasina=5/13cosa=12/13所以cos(3π+a)=-12/13
属于(π/2,π),是吧?所以sinb=2√2/3,所以2√2/3cosa-1/3sina=7/9,所以sina=2√2/3cosa-7/3所以(2√2/3cosa-7/3)ˆ2+cosa&
a∈(0,π/2)时,tana-cota=sina/cosa-cosa/sina=(sina+cosa)(sina-cosa)/(sinacosa)因为sina>0,cosa>0,若使cota0,这与
sin(a+π/2)=cosa=1/3因为,a属于(-π/2,0)所以sina
由sinA-cosA=1得sin(a-b/2)=九分之四倍根号5,cos(a/2-b)=3分之根号5由cos(a-b/2)*cos(a/2-b)-sin(a-b/2)*sin(a/2-b)=cos(a
(1)由m•n=12,得cos2A2−sin2A2=12,即cosA=12∵A为△ABC的内角,∴A=π3(2)由余弦定理:a2=b2+c2-2bccosA⇒a2=(b+c)2-3bc即12=42-3
(Ⅰ)∵sinA2=55,0<A<π∴cosA2=255.∴sinA=2sinA2cosA2=45.∵S△ABC=12bcsinA=2,∴bc=5.(Ⅱ)∵sinA2=55,∴cosA=1−2sin2
tana=tan(a-b+b)=[tan(a-b)+tanb]/[1-tan(a-b)*tanb]=(1/2-1/7)/(1+1/2*1/7)=1/3所以tan(2a-b)=tan(a+a-b)=[t
(1)f(A)=2cos2AsinA2cosA+cos2A+12(2分)=cosA•sinA+cos2A+12(1分)=12(sin2A+cos2A+1)(1分)=22sin(2A+π4)+12(2分
sinA=sin((A+B)-B)=sin(A+B)cosB-sinBcos(A+B)由题意知:A+B∈(π/2,3π/2)∴cos(A+B)=-4√2/9sinB=2√2/3之后再代入上式就好解多了
f(x)=(sinx)²-sinx-a=(sinx-0.5)²-a-0.5,x∈[0,2π]∵sinx∈[-1,1]∴f(x)在sinx=0.5时取得最小值-a-0.5,f(x)在
能把题写的清楚一点吗,这样没法求解.
sinA2-sinB2=(sinA+sinB)(sinA-sinB)=(sinA+sinB)(sinA-sinB)=2sin((A+B)/2)cos((A-B)/2)*2Cos((A+B)/2)Sin
因为a属于(0,π/2),所以0
1、a不等于0p=5*6*6/(6*6*6)=5/62、a=0,b不为0p=1*5*6/(6*6*6)=5/36