化简:❶sin(2a+b)/sina-2cos(a+b)=sinb/sina ❷cos4a+
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化简:❶sin(2a+b)/sina-2cos(a+b)=sinb/sina ❷cos4a+4cos2a+3=8cos⁴a
![化简:❶sin(2a+b)/sina-2cos(a+b)=sinb/sina ❷cos4a+](/uploads/image/z/16194520-64-0.jpg?t=%E5%8C%96%E7%AE%80%EF%BC%9A%26%2310102%3Bsin%EF%BC%882a%2Bb%EF%BC%89%2Fsina-2cos%28a%2Bb%29%3Dsinb%2Fsina+%26%2310103%3Bcos4a%2B)
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
=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?