解答
展开 (A+(C⋅cos(y))+(D⋅cos(z))−(B⋅cos(x)))2
解答
A2+C2cos2(y)+D2cos2(z)+B2cos2(x)+2ACcos(y)+2ADcos(z)−2ABcos(x)+CDcos(y)cos(z)−CBcos(y)cos(x)+CDcos(z)cos(y)−DBcos(z)cos(x)−CBcos(x)cos(y)−DBcos(x)cos(z)
求解步骤
(A+(Ccos(y))+(Dcos(z))−(Bcos(x)))2
化简 A+(Ccos(y))+(Dcos(z))−(Bcos(x)):A+Ccos(y)+Dcos(z)−Bcos(x)
=(A+Ccos(y)+Dcos(z)−Bcos(x))2
使用指数法则: ab+c=ab⋅ac(A+Ccos(y)+Dcos(z)−Bcos(x))2=(A+Ccos(y)+Dcos(z)−Bcos(x))(A+Ccos(y)+Dcos(z)−Bcos(x))=(A+Ccos(y)+Dcos(z)−Bcos(x))(A+Ccos(y)+Dcos(z)−Bcos(x))
打开括号=AA+ACcos(y)+ADcos(z)+A(−Bcos(x))+Ccos(y)A+Ccos(y)Ccos(y)+Ccos(y)Dcos(z)+Ccos(y)(−Bcos(x))+Dcos(z)A+Dcos(z)Ccos(y)+Dcos(z)Dcos(z)+Dcos(z)(−Bcos(x))−Bcos(x)A−Bcos(x)Ccos(y)−Bcos(x)Dcos(z)−Bcos(x)(−Bcos(x))
化简 AA+ACcos(y)+ADcos(z)+A(−Bcos(x))+Ccos(y)A+Ccos(y)Ccos(y)+Ccos(y)Dcos(z)+Ccos(y)(−Bcos(x))+Dcos(z)A+Dcos(z)Ccos(y)+Dcos(z)Dcos(z)+Dcos(z)(−Bcos(x))−Bcos(x)A−Bcos(x)Ccos(y)−Bcos(x)Dcos(z)−Bcos(x)(−Bcos(x)):A2+C2cos2(y)+D2cos2(z)+B2cos2(x)+2ACcos(y)+2ADcos(z)−2ABcos(x)+CDcos(y)cos(z)−CBcos(y)cos(x)+CDcos(z)cos(y)−DBcos(z)cos(x)−CBcos(x)cos(y)−DBcos(x)cos(z)
=A2+C2cos2(y)+D2cos2(z)+B2cos2(x)+2ACcos(y)+2ADcos(z)−2ABcos(x)+CDcos(y)cos(z)−CBcos(y)cos(x)+CDcos(z)cos(y)−DBcos(z)cos(x)−CBcos(x)cos(y)−DBcos(x)cos(z)