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求证:sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα.
sin(2a+b)/sina-2cos(a+b) =sin[(a+b)+a]/sina-2cos(a+b) =[sin(a+b)cosa+cos(a+b)sina]/sina-2cos(a+b) =[sin(a+b)cosa/sina+cos(a+b)]-2cos(a+b) =sin(a+b)cosa/sina-cos(a+b) =[sin(a+b)cosa-cos(a+b)sina]/sina =sin[(a+b)-a]/sina =sinb/sina.证完。
[sin(2α+β)/sinα-2cos(α+β)]*sina =sin(2α+β)-2cosαcosβsinα+2(sinα)^2*sinβ =sin(2α+β)-sin2αclsβ+sinβ-sinβcos2α =sin(2α+β)-sin(2α+β)+sinβ=sinβ 所以sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
答:a·b=2cosA*3cosB+2sinA*3sinB=6(cosAcosB+sinAsinB)=6cos(A-B) a·b=|a|*|b|cos60=2*3(...详情>>
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