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;