(3-4cos 2A+cos 4A)/(3+4cos 2A+cos 4A)化简,跪求化简啊啊啊啊~~~~~~~~~~~急
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(3-4cos 2A+cos 4A)/(3+4cos 2A+cos 4A)化简,跪求化简啊啊啊啊~~~~~~~~~~~急
(3-4cos 2A+cos 4A)/(3+4cos 2A+cos 4A)化简,跪求
化简啊啊啊啊~~~~~~~~~~~急
(3-4cos 2A+cos 4A)/(3+4cos 2A+cos 4A)化简,跪求化简啊啊啊啊~~~~~~~~~~~急
(3-4cos 2A+cos 4A)/(3+4cos 2A+cos 4A)
=(2cos 2A^2-4cos 2A+2)/(2cos 2A^2+4cos 2A+2)
=(cos 2A-1)^2/(cos 2A+1)^2
=tanA^4
cos^2A+cos^2(2π/3+A)+cos^2(4π/3+A)
由cos(a+b)=cos a cos b-sin a sin b cos(a-b)=cos a cos b+sin a sin b解题设a为锐角,证:1、2分之根3乘cos a + 2分之1乘sin a=cos(6分之π-a)2、cos a-sin a=根号2cos(4分之π+a)
5cos^2a+4cos^2β=4cosα则cos^2a+cos^2β
三角函数 求证:cos^2(a)-cos(2a)*cos(4a)=sin^2(3a)求证:cos^2(a)-cos(2a)*cos(4a)=sin^2(3a)
化简cos^2 a(2cos^2 a+3)-sin^2 a(2cos^2 a+3)-4sin^2 a cos^2 a +3
化简 1-sin^4a-cos^4a/cos^2a-cos^4a
化简(1-sin^6 a-cos^6 a)/(cos^2 a-cos^4 a)==
已知sin(3π+a)=1/4,求[cos(π+a)]/{cosa[cos(π+a)-1]}+[cos(a-2π)]/[cos(a+2π)cos(π+a)+cos(-a)]的值要具体过程
证明(3-4cos 2A+cos 4A) / (3+4cos 2A+cos 4A)=tan^4 A
tan a =3求4sin a-2cos a/5cos+3sin a
化简cos^4 a-(1/4)cos^2 2a-(1/2)cos 2a
cos (2 a/7)+cos(4 a/7)+cos(6 a/7)=?
(3-4cos 2A+cos 4A)/(3+4cos 2A+cos 4A)化简,跪求化简啊啊啊啊~~~~~~~~~~~急
求证1/sin^2a+3/cos^2a>=4+2根号下31/sin²a +3/cos²a=(sin²a+cos²a)/sin²a +3(sin²a+cos²a)/cos²a=1+cos²a/sin²a+3+3sin²a/cos²a=4+cos²a/sin²a+3sin²a/cos²a≥4+2√[(cos
sin^4+sin^2cos^2+cos^2a=
COS(4a)+4COS(2a)=[8(COSa)^4]-3求教如何证明
已知cos(a+b)=4/5,cos(a-b)=-4/5,3π/2
.已知Cos(a+b)=4/5,cos(a-b)= -4/5,3/2π