Question

Докажите тождество!!! sin2a(sin2a+sin2b) +cos2a(cos2a+cos2b)=2cos(в квадрате) (a-b)

Answer 1

sin2a(sin2a+sin2b) +cos2a(cos2a+cos2b)=sin²2a + sin2asin2b + cos²2a + cos2acos2b=(sin²2a+cos²2a)+sin2asin2b +cos2acos2b=1+cos(2a-2b)=1+cos2(a-b)=

=1+cos²(a-b) - sin²(a-b)=2cos²(a-b)