Question

Выразите sin56° через а, если tg17°=a

Answer 1

sin(56°) = sin(90° - 34°) = cos(34°) = cos(2*17°) = cos^2(17°) - sin^2(17°) =
= ( cos^2(17°) - sin^2(17°) )/( cos^2(17°) + sin^2(17°) ) = V
Разделим числитель и знаменатель последней дроби на cos^2(17°).
V = (1 - [sin^2(17°)/cos^2(17°)] )/( 1 + [sin^2(17°)/cos^2(17°)] ) =
= (1 - tg^2(17°) )/( 1 + tg^2(17°) ) = (1- a^2)/(1+ a^2).

Answer 2

sin56=cos34=1-tg²17/1+tg²17=1-a²/1+a²→ответ.