Ответы и объяснения

2013-03-28T21:55:21+04:00

(1-cos2α+sin2α)/(sinα+cosα)=(2sin²α+2sinαcosα)/(sinα+cosα)=2sinα(sinα+cosα)/(sinα+cosα)= 2sinα

2013-03-28T22:03:42+04:00

1-cos2a+sin2a/cosa+sina=

cos^2a+sin^2a-cos2a+sin2a/cosa+sina

 sin^2a-cos^2a+2cosa*sina+1/cosa+sina=

 sin^2a-cos^2a+sin2a+1/cosa+sina=2sina

 cos2a(1+tg^2a)-1=

tg^2a= sin^2/cos^2a

cos^2-sin^2a

1-2sina^2a+ (1-2sina^2*sin^2a/cos^2a)-1=-2sin^2a+1-2sin^4a/cos^2a=-sin^2a-2sin^2a/cos^2a=-tg^2a