(sinasinb)^2+(cosacosb)^2-(1/2)(cos2acos2b)求值(过程..)
问题描述：
(sinasinb)^2+(cosacosb)^2-(1/2)(cos2acos2b)求值(过程..)
答
(sinasinb)^2+(cosacosb)^2-(1/2)(cos2acos2b)
=(1/4)(1-cos2a)(1-cos2b)+(1/4)(1+cos2a)(1+cos2b)-(1/2)cos2acos2b
=1/2+(1/2)cos2acos2b-(1/2)cos2acos2b
=1/2
答
=(1-cosa^2)(1-cosb^2)+(cosacosb)^2-(1/2)(cos2acos2b)
=1-cosa^2-cosb^2+2cosa^2 cosb^2-(1/2)(cos2acos2b)
=1-(1/2)(cos2a+1)-(1/2)(cos2b+1)-(1/2)(cos2acos2b)
括号展开，化简得；=1/2
答
sina=x sinb=y
cosa=m cosb=n
cos2a=P cos2b=Q
P=(cosa)^2-(sina)^2=y^2-x^2 Q=n^2-m^2
原式=xxyy-mmnn-1/2(PQ)
=xxyy-mmnn-1/2(yynn+mmxx-yyxx-mmnn)
=1/2(yyxx+mmnn-yynn-mmxx)
=1/2(yy-mm)(xx-nn)
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