Докажите что если tga=1/7, и tgb=1/3 то cos2a=sin4b

Tgα=1/7; tgβ=1/3; ⇒cos2α=sin4β;

cos2α = (1-tg²α) / (1+tg²α) = (1-1/49) / (1+1/49) = 48/50;

sin4β=2sin2β·cosβ;

sin2β=2tgβ / (1+tg²β) = 2· (1/3) / (1+1/9) = 2/3:10/9=3/5;

cos2β = (1-tg²β) / (1+tg²β) = (1-1/9) / (1+1/9) = 8/10;

sin4β=2·3/5·8/10=48/50;

48/50=48/50⇒cos2α=sin4β;