二倍角的三角函数sin²asin²p+cos²acos²p=(1/2)(1+co

把左式的平方项化成二倍角：
sin^2a=1/2(1-cos2a)
sin^2p=1/2(1-cos2p);
cos^2a=1/2(1+cos2a)
cos^2p=1/2(1+cos2p)
左式=1/4[(1-cos2a)(1-cos2p)+(1+cos2a)(1+cos2p)]
=1/4[（1+cos2acos2p)-(cos2a+cos2p)+(1+cos2acos2p)+(cos2a+cos2p)]
=1/2(1+cos2acos2p)
左式=右式 即为所证
对不起,我看见你改过的题晚了一些时间.以上供你参考!