Question

Sin2x+cos2x=0 решите подробно

Answer 1

sin2x  =  2tgx/(1 + tg2x);
cos2x  =  (1 - tg2x)/(1 + tg2x);
Sin2x+cos2x= 2tgx/(1 + tg2x)+(1 - tg2x)/(1 + tg2x)=(1 - tg2x+2tgx)/(1 + tg2x);
tg2x=2tgx/(1-tg^2 x);
1 + tg2x= (1-tg^2 x)/(1-tg^2 x)+2tgx/(1-tg^2 x)=-(tgx-1)^2/(tgx-1)(tgx+1)=(1-tgx)/(1+tgx);
1 - tg2x+2tgx=1+2tgx-2tgx/(1-tgx)(tgx+1)=((1-tgx)(tgx+1)+2tgx(1-tgx)(tgx+1)-2tgx)/(1-tgx)(tgx+1);
((1-tgx)(tgx+1)+2tgx(1-tgx)(tgx+1)-2tgx)/(1-tgx)^2;
(tg^2 x - 2tgx +1)/(1-tgx)^2;
(tgx-1)^2/(1-tgx)^2=1 тк (1-tgx)^2=(tgx-1)^2