Question

Se sen x = m e cos x = n, determiné o sen 2x, cos 2x e tg2x.

Answer 1

a) sen2x = 2senx.cosx = 2mn

b) cos2x = cos²x - sen²x = n² - m²

tgx = senx/cosx = m/n

$tg2x= \frac{2tgx}{1-tg^2x} = \frac{2. \frac{m}{n} }{1- \frac{m^2}{n^2} } = \frac{ \frac{2m}{n} }{ \frac{n^2-m^2}{n^2} } = \frac{2m}{n} * \frac{n^2}{n^2-m^2} = \frac{2mn}{n^2-m^2}$