爲什麽1/2(sin2A sin2B)=sin(A B)cos(A-B) 網友廻答:
- 1/2(sin2A sin2B)=1/2(sin(A B A-B) sin((A B)-(A-B)))=1/2(sin(A B)cos(A-B) sin(A-B)cos(A B) sin(A B)cos(A-B)-sin(A-B)cos(A B))=1/2*2sin(A B)cos(A-B)=sin(A B)cos(A-B)
網友廻答:
- 左邊展開爲1/2(2sinAcosA 2sinBcosB)=sinAcosA sinBcosB
右邊展開爲(sinAcosB cosAsinB)(cosAcosB sinAsinB)=sinAcosA(cosB)^2 sinAcosA(sinB)^2 sinBcosB(cosA)^2 sinBcosB(sinA)^2=sinAcosA((cosB)^2 (sinB)^2) sinBcosB((sinA)^2 (cosA)^2)=sinAcosA sinBcosB
兩式相等
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