cos2A+cos(4π/3-2A)化简,
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cos2A+cos(4π/3-2A)化简,
cos2A+cos(4π/3-2A)化简,
cos2A+cos(4π/3-2A)化简,
=cos2A-cos(π/3-2A)=cos2A-1/2cos2A-2分之根号3sin2A=1/2cos2A-2分之根号3sin2A=cos(2A+π/3)
cos2A+cos(4π/3-2A)化简,
化简cos^2(π/4-a)-sin^2(π/4-a)得到?A.sin2a B.-sin2a C.cos2a D.-cos2a
若cos(π/4-a)cos(π/4+a)=(根号2 )/6,则cos2a
化简[2sin2a/(1+cos2a)]*(cos^2a/cos2a)
为什么cos(π+2a)等于-cos2a
已知sin a+cos a=1/3,a属于(π/2,π),求cos2a
化简:tan(π/4+a)cos2a / 2cos^2(π/4-a) 在线等
cos(a-b)=-4/5,cos(a+b)=4/5 a属于(π/2,π)b属于(3π/2,2π),求cos2a
已知cos(a+?π)=3/5,π/2≤a<3π/2.求cos2a.更改:已知cos(a+π/4)=3/5
化简:❶sin(2a+b)/sina-2cos(a+b)=sinb/sina ❷cos4a+4cos2a+3=8cos⁴a
已知cos[(3π+a)/2]=-2/3则cos2a=
在三角形ABC中,cos(π/4+A)=3/5,则cos2A的值
在三角形ABC中,已知cos(π/4+A)=3/5,则cos2A=
化简:{cos(4a-兀/2)乘以 sin(兀/2-a) 乘以 cos2a } / { (1+cos2a) 乘以 (1+cos4a)乘以cosa}
1-cos2a=?cos(-π-a)=?cos(3π/2-a)=?sin^2a-cos^2a=?
化简[cos(4a-π/2)*sin(π/2-a)*cos2a]/[(1+cos2a)(1+cos4a)*cosa知道很麻烦,但是还是希望有那位好心人帮我解决一下,
已知a属于(0,二分之π),且2sin^2a -sinacosa-3cos^2a=0,求sin(a+π/4)/(sin2a+cos2a+1)
化简sin^4a-cos^4a/cos2a