化简:(1)(cos²10°-cos²80°)²+cos²70°=? (2)(sinα-cosα)²+sin4α/2cos2α=
问题描述:
化简:(1)(cos²10°-cos²80°)²+cos²70°=? (2)(sinα-cosα)²+sin4α/2cos2α=
答
(1)=(sin280-cos280)2+cos270
=cos2160+cos270
=sin270+cos270
=1 (2为平方,不好意思,
(2)=1-sin2a+2sin2acos2a/2cos2a
=1-sin2a+sin2a
=1