Question

Срочно!
Найти sin 2a,cos 2a,tg 2a, если sin a=1/3 П/2<а<П

Answer 1

cos a = $sqrt{1-sin^2a}$ = 2$sqrt{2}$/3
sin 2a = 2sin a cos a = 2*1/3*2$sqrt{2}$/3 = 2$sqrt{2}$/3

tg a = sina/cos a = 1/$2 sqrt{2}$
tg 2a = 2tga/1-tg^2a = (1/$sqrt{2}$) / 1 - 1/8 = 8/7$sqrt{2}$