Ответы и объяснения

2013-10-30T10:57:43+00:00
=cos²α+sin²α+2sinαcosα+cos²α-sin²α==2c0s²α+2sinαc0sα =2cosα(cosα+sinα)==2cosα(sin(π/2-α)+sinα)=2cosα·2sin(π/2-α+α)/2·cos(π/2-α-α)/2= =4cosα·sinπ/4·cos(π/4-α)=4cosα·√2/2·cos(π/4-α)=2√2cosα·cos(π/4-α)