Question

Sendo sen x= 3/5 e sen y= 13/15 no 1 quadrante, calcule:
A) Sen (x+y)
B) Tg (x+y)

Answer 1

Utilizando as propriedades trigonométricas, temos que:

sen²(x) + cos²(x) = 1

(3/5)² + cos²(x) = 1

cos²(x) = 1 - 9/25 = 16/25

cos(x) = 4/5

(13/15)² + cos²(y) = 1

cos²(y) = 1 - 169/225 = 56/225

cos(y) = 2√14/15

Agora, fazemos:

a) sen(x+y) = sen(x)cos(y) + sen(y)cos(x)

sen(x+y) = (3/5)*(4/5) + (13/15)*(2√14/15)

sen(x+y) = (12/25) + (26√14/225)

sen(x+y) = (108+26√14)/225

b) tg(x+y) = tg(x).tg(y)/1-tg(x).tg(y)

Como tg(x) = sen(x)/cos(x) e tg(y) = sen(y)/cos(y):

tg(x+y) = [sen(x)sen(y)/cos(x)cos(y)]/[1 - sen(x)sen(y)/cos(x)cos(y)]

tg(x+y) = [3/5 . 13/15 / 4/5 . 2√14/15]/[1 - sen(x)sen(y)/cos(x)cos(y)]

tg(x+y) = (39/75)/(8√14/75)/[1 - sen(x)sen(y)/cos(x)cos(y)]

tg(x+y) = (39/8√14)/[1 - sen(x)sen(y)/cos(x)cos(y)]

tg(x+y) = (39√14/112)/[1 - 39√14/112]