化简：❶sin（2a+b）/sina-2cos(a+b)=sinb/sina ❷cos4a+

来源：学生作业帮编辑：搜狗做题网作业帮分类：数学作业时间：2024/08/18 15:22:31

化简：❶sin（2a+b）/sina-2cos(a+b)=sinb/sina ❷cos4a+4cos2a+3=8cos⁴a

sin（2a+b）/sina-2cos(a+b)
=sina*cos(a+b)/sina+sin(a+b)cosa/sina-2cos(a+b)*sina/sina
=[sin(a+b)cosa-cos(a+b)sina]/sina
=sin(a+b-a)/sina
=sinb/sina
cos4a+4cos2a+3
=2(cos2a)^2-1+4cos2a+3
=2[(cos2a)^2+2cos2a+1]
=2[cos2a+1]^2
=2[2(cosa)^2-1+1]^2
=8(cosa)^4

为什么sinA-sinB/sinA+sinB=cos[(A+B)/2]sin[(A-B)/2]/{sin[(A+B)/2 证明sin(2a+b)/sina-2cos(a+b)=sinb/sina 求证sin(2A+B)/sinA-2cos(A+B)=sinB/sinA sinb/sina=cos(a+b),证明3sinb=sin(2a+b) 为什么sina+sinb==2sin(a+b)\/2*cos(a-b)\/2 为什么sina+sinb==2sin(a+b)/2*cos(a-b)/2 三角形ABC中,为什么sinA+sinB=2sin(A+B)/2*cos（A-B)/2 求证sina+sinb=2sin(a+b)/2*cos(a-b)/2 如何证明sinA+sinB=2sin((A+B)/2)cos((A-B)/2) sinA+sinB=2sin((A+B)/2)cos((A-B)/2 sina+sinb=2sin((a+b)/2)cos((a-b)/2的推导过程 sin(A+B)-sinA=2cos(A+B/2)*sinB/2怎么推导?