三角恒等变换的化简
问题描述：
三角恒等变换的化简
sin²A+sin²B-sin²Acos²B-2sinAcosBcosAsinB-cos²Asin²B=sinAsinB
2sin²Asin²B-2sinAcosBsinBcosA=sinAsinB
怎么化简的?
接着上面的cosAcosB-sinAsinB=-1/2
cos(A+B)=-1/2
怎么化简的？
答
sin²A+sin²B-sin²Acos²B-2sinAcosBcosAsinB-cos²Asin²B
=(sin²A-sin²Acos²B)-2sinAcosBcosAsinB+(sin²B-cos²Asin²B)
=sin²A(1-cos²B)-2sinAcosBcosAsinB+sin²B(1-cos²A)
=sin²Asin²B-2sinAcosBcosAsinB+sin²Bsin²A
=2sin²Asin²B-2sinAcosBsinBcosA
=2sinAsinB(sinAsinB-cosBcosA)
=2sinAsinB[-cos(B+A)]
=2sinAsinBcosC
∵sin²A+sin²B-sin²Acos²B-2sinAcosBcosAsinB-cos²Asin²B=sinAsinB
∴2sinAsinBcosC=sinAsinB
∴cosC=1/2,
∴C=60°