Ответы на вопрос:
1/2*(сos(5a-4a)+cos(5a+/2*(cos(4a-3a)+cos(4a+3a))+2cos²2acosa= =-1/2*cosa-1/2*cos9a-1/2*cosa-1/2*cos7a+2cos²2acosa= =-cosa+2cos²acosa-1/2(cos9a+cos7a)=cosa(-1+2cos²2a)-1/2*2cos8acosa= =-cosa*cos4a-cosacos8a=-cosa*(cos4a+cos8a)=-cosa*2cos6acos2a= =-2cosacos2acos6a
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